IN  MEMORIAM 
FLOR1AN  CAJORI 


PLANE  AND  SOLID 


GKOMBTRY 


SUGGESTIVE  METHOD 


BY 


C.  A.  VAN  VELZER 
Professor  of  Mathematics,  University  of  Wisconsin, 


AND 


GEO.  C.  SHUTTS 

Professor  of  Mathematics  and  History,  State  Normal  School, 
Whitewater,  Wis. 


MADISON,  WIS. 
TRACY,  GIBBS  &  CO. 


COPYRIGHT,    1894. 

C.  A.  VAN  VELZER  AND  GEO.  C.  SHUTTS. 


Tracy,  Gibbs  &  Co.,  Printers  and  Stereotypers. 


PREFACE. 


An  examination  will  show  that  this  book  is  constructed  upon  a 
unique  plan,  and  to  give  proper  credit  for  this  plan  I  have  requested 
the  privilege  of  writing  this  preface  myself.  The  plan  is  due  entirely 
to  Professor  Shutts  who  has  used  it  for  years  from  mimeograph  notes 
with  such  excellent  results  that  it  was  thought  best  to  put  the  matter 
in  permanent  form  for  a  text-book.  Accordingly  the  mimeograph  notes 
were  carefully  revised  by  Professor  Shutts  and  myself  and  put  into 
nearly  the  present  form,  when  the  work  was  again  tested  by  Professor 
Shutts  in  his  classes  in  the  State  Normal  School  at  Whitewater,  Wis. 
After  this  further  trial  the  work  was  again  revised  and  improved. 
The  method  of  this  book  therefore  is  not  untried,  but  has  been  sub- 
jected to  a  thorough  test  and  found  to  give  most  excellent  results. 

In  putting  the  work  into  its  present  form  the  scientific  classifica- 
tion of  the  subject-matter  has  been  departed  from  when  it  was 
thought  that  by  so  doing  the  work  could  be  better  graded  to  the  abil- 
ity of  the  average  pupil.  For  this  reason  the  subject  of  triangles  has 
been  introduced  before  the  relations  of  lines  and  angles  have  been 
fully  discussed. 

A  departure  from  ordinary  methods  will  be  noticed  in  the  treat- 
ment of  proportion.  It  has  not  been  thought  wise  to  follow  the  or- 
dinary method  in  this  country  of  limiting  the  subject  to  proportions 
whose  terms  are  pure  numbers,  nor  yet  to  follow  the  Euclidian 
method  common  in  England,  which  indeed  admits  of  proportions 
whose  terms  are  concrete  magnitudes,  but  which  is  so  difficult  that  it 
can  be  understood  by  only  the  best  students.  The  method  in  the 
text  will  be  found  to  admit  of  proportions  whose  terms  are  concrete 
magnitudes  and  yet  will  do  no  violence  to  the  fundamental  ideas  of 
arithmetic  regarding  operations  upon  concrete  magnitudes. 


M306168 


IV  PREFACE. 


The  subject  of  limits  is  treated  in  a  manner  so  simple  that  begin- 
ners can  grasp  it. 

With  an  average  high  school  class  all  the  work,  including  exercises, 
can  be  done  in  one  year. 

In  preparing  this  book  for  the  press  the  works  which  have  been 
consulted  most  freely  and  have  been  most  serviceable  are  Byerly's 
edition  of  Chauvenet's  Geometry,  and  the  geometry  prepared  by  the 
English  Society  for  the  Improvement  of  Geometrical  Teaching. 

The  thanks  of  the  authors  are  due  to  Professor  A.  J.  Hutton,  of  the 
State  Normal  School,  of  Platteville,  Wis.,  for  valuable  suggestions 
and  criticisms,  C.  A.  VAN  VELZER. 


SUGGESTIONS  TO  TEACHERS. 


Geometry  is  essentially  a  disciplinary  study,  and  benefit  derived 
from  its  study  depends  upon  the  independent  thought  expended  by 
the  pupil.  A  geometry  is  in  the  nature  of  a  "key"  to  the  extent  to 
which  the  demonstrations  are  written  out  for  the  pupil.  That  part 
of  the  work  which  a  pupil  can  do  for  himself  should  not  be  done  for 
him.  The  teacher  and  text-book  should  furnish  the  data  and  stimu- 
late thought  rather  than  give  him  a  set  form  of  words  which  he  may 
repeat  entire  with  or  without  the  ideas  which  those  words  should 
convey. 

In  the  following  pages  are  suggestions  arranged  in  logical  order 
which  are  intended  to  direct  and  stimulate  the  thought  of  the  pupil 
so  that  he  may  work  out  his  own  demonstrations. 

Model  demonstrations  are  given  of  a  few  propositions  to  show  the 
student  how  to  work  out  his  own  demonstrations,  and  in  what  form 
they  should  be  given.  By  comparison  it  will  be  seen  that  the  answers 
to  the  suggestions  logically  linked  together,  constitute  the  demonstra- 
tion. The  suggestions  should  be  studied  in  their  order,  for  usually 
each  suggestion  depends  upon  the  preceding  one.  The  answer  to  a 
suggestion  should  consist  of  a  statement  of  the  relations  asked  for, 
together  with  the  authority  in  full  for  such  statement.  To  be  indif- 
ferent in  regard  to  the  authority  in  any  instance  is  to  encourage  care- 
lessness, slovenliness  and  inaccuracy  in  demonstration.  A  common 
error  is  to  apply  authority  that  does  not  exactly  fit  the  conditions  un- 
der consideration.  The  pupil  should  be  made  to  clearly  understand 
that  the  authority  should,  without  exception,  be  a  definition,  an 
axiom,  or  a  previously  proved  proposition.  "It  seems  so,"  or,  "it 
looks  reasonable,"  or  any  expression  of  judgment  will  not  do.  The 


vi  SUGGESTIONS   TO   TEACHERS. 

pupil  should  be  encouraged  to  search  out  his  own  authority,  even 
when  the  authority  is  quoted  for  him  in  the  suggestions,  using  the 
reference  simply  for  verification.  A  pride  in  independent  work  is  a 
most  important  factor  in  securing  satisfactory  results. 

In  the  preparation  of  the  lesson  the  pupil  should  write  out  his 
demonstration,  noting  carefully  the  form  of  the  "models."  This 
will  ensure  correct  form  and  avoid  haziness  of  thought.  During  the 
first  few  weeks  the  teacher  should  scan  carefully  this  written  work 
as  well  as  tests  taken  in  the  recitation. 

The  exercises,  or  at  least  a  part  of  them,  should  be  taken  along 
with  the  propositions  as  they  occur,  and  not  be  studied  all  together 
at  the  end  of  a  chapter. 

Time  will  be  saved  in  the  end  by  starting  slowly  but  surely,  pass- 
ing over  nothing  that  is  not  clearly  understood. 

Since  each  demonstration  involves  previous  propositions  and  def- 
initions, facility  in  demonstration  can  be  best  secured  by  committing 
to  memory  each  theorem  and  definition;  for  that  authority  cannot  be 
readily  recognized  and  applied  which  is  only  imperfectly  in  mind. 


CONTENTS. 


PLANE   GEOMETRY. 

CHAPTER  I. 

PAOE. 

RECTILINEAR  FIGURES,         ------        .-1 

PROPOSITIONS  IN  CHAPTER  I,-        ------      77 

CHAPTER  II. 

THE  CIRCLE, ...-82 

PROPOSITIONS  IN  CHAPTER  II,      -------    133 

CHAPTER  III. 

PROPORTIONAL  LINES  AND  SIMILAR  POLYGONS,    -  137 

PROPOSITIONS  IN  CHAPTER  III,    -  ....    173 

CHAPTER   IV. 

COMPARISON  AND  MEASUREMENT  OF  POLYGONS,          ...     17G 
PROPOSITIONS  IN  CHAPTER  IV,    -  -        ....    206 

CHAPTER  V. 

REGULAR  POLYGONS  AND  CIRCLES,     --....    209 
PROPOSITIONS  IN  CHAPTER  V,     -•-....    239 


Vlll  CONTENTS. 


GEOMETRY  OF  THREE  DIMENSIONS. 

CHAPTER  VI. 

PAGE. 

LINES  AND  PLANES,     -  --241 

PROPOSITIONS  IN  CHAPTER  VI,    -------    283 

CHAPTER  VII. 

POLYHEDRONS, 287 

PROPOSITIONS  IN  CHAPTER  VII,  ......    335 

CHAPTER  VIII. 


THE  THREE. ROUND  BODIES, 
PROPOSITIONS  IN  CHAPTER  VIII, 

INDEX,  • 


CHAPTER  I. 
RECTILINEAR  FIGURES. 


DEFINITIONS. 

1.  The  block  represented  in  the  figure  occupies  a  lim- 
ited portion  of  space.     Conceive  it  to  be  removed.     Its 
form  or  shape  may  still   be   retained 

in  mind.     This  is  true  of  any  object 
or  body. 

The  space  conceived  to  be  occupied 
by  a  body,  apart  from  its  substance,  is 
called  a  geometrical  solid.  The 
matter  or  substance  of  which  the 
body  is  composed  constitutes  a  physical  solid.  Hence 
a  geometrical  solid  is  simply  the  shape  or  form  of  a 
physical  solid,  or  some  form  or  figure  conceived  by 
the  mind. 

Definition.  A  geometrical  solid  is  a  limited  portion 
of  space  and  has  length,  breadth  and  thickness. 

The  terra  solid  will  hereafter  be  used  for  geometrical  solid. 

2.  The  boundary  that  separates  a  solid  from  the  space 
outside  of  itself  is  called  the  surface  of  the  solid.     Dis- 
tinct  flat  portions   of  the  bounding  surface  are   called 
faces. 

As  we  can  think  of  the  surface  of  a  body  without  in- 
cluding any  of  its  substance,  surface  may  be  regarded  as 
having  no  thickness. 

Definition.  Surface    is   that   which  has   length  and 
breadth  without  thickness. 
1— Geo. 


2  PLANE   GEOMETRY. 

3.  If  a  surface  be  divided  into  distinct  portions,  the 
boundaries  of  those  portions  are  called  lines.     In  the 
solid  represented  on  page  1,  the  edges,  or  boundaries  of 
the  faces,  are  lines.     These  lines  being  the  intersections 
of  faces  which  have  no  thickness  can  themselves  have 
neither  breadth  nor  thickness. 

Definition.  A  line  is  that  which  has  length  but  has 
neither  breadth  nor  thickness. 

4.  If  a  line  be  divided  into  distinct  portions  the  limits 
of  these  portions  are  called  points.     In  the  solid  repre- 
sented on  page  1,  the  corners,  or  limits  of  the  edges,  are 
points.     These   points  being  the   intersections    of  lines 
which  have  neither  breadth  nor  thickness  can  themselves 
have  neither  length,  breadth  nor  thickness. 

Definition.  A  point  is  that  which  has  simply  posi- 
tion but  neither  length,  breadth  nor  thickness. 

5.  A  surface  may  be  conceived  apart  from  a  solid,  a 
line  apart  from  a  surface  and  a  point  apart  from  a  line. 
If  a  point  be  conceived  to  move,  the  path  in  which  it 
moves  is  a  line,  hence  a  line  may  be  regarded  as  the 
path,  or  locus,  of  a  moving  point. 

When  a  line  is  regarded  as  formed  by  a  moving  point 
the  point  is  said  to  describe  the  line. 

6.  A  geometrical  figure  is  a  combination  of  points, 
lines,  surfaces  or  solids  in  any  given  relations. 

Geometrical  figures  are  ideal,  i.  e.,  mental  conceptions, 
though  none  the  less  real,  and  can  be  represented  to  the 
eye  only  by  some  material  substance;  for  instance,  a  line 
may  be  represented  by  a  mark  made  by  a  pencil,  crayon, 
etc.  A  solid  may  be  represented  by  a  drawing  or  a 
block  of  wood  or  some  material  in  any  given  shape. 


DEFINITIONS. 


To  avoid  multiplying  words,  the  representation  of 
geometrical  figures  will  generally  be  referred  to  for  the 
concepts  themselves. 

7.  A  straight  line  is  a  line  such  that  any  part  of  it, 
however  placed,  will  lie  wholly  in  any  other  part,  if  its 
extremities  be  made  to  lie  in  that  part.     A  line  is  read 
by  naming  the  letters  placed  at  its  extremities,  as  line 
AB.  ± 2. 

8.  A  broken  line   is  one  made 
up    of    a    succession    of    different 
straight  lines,  as  A  'B  CD  E. 

9.  A  curved  line,  or  a  curve,  is 
a    line    no    portion    of    which    is 
straight,  as  C  D. 

10.  A  plane  surface,  or  a  plane, 

is  a  surface  such  that  if  any  two  of  its  points  be  joined 
by  a  straight  line  the  line  will  lie  wholly  in  the  surface. 

11.  A  plane  figure  is  a  figure  which  lies  wholly  in 
the  same  plane. 

12.  A  plane  figure  which  is  made  wholly  of  straight 
lines  is  called  a  rectilinear  figure. 

13.  When  reference  is  made  simply  to  extent,  figures 
are  called  magnitudes. 

14.  Geometry  is  the  science  which  treats  of  points, 
lines,  surfaces  and  solids,  and  is  concerned  with  the  con- 
struction and  measurement  of  geometric  figures. 

15.  Plane  geometry  treats  of  plane  figures. 


PLANE    GEOMETRY. 


16.  Solid  geometry  treats  of  figures  which  are  not 
wholly  in  the  same  plane. 

ANGLES. 

17.  When  two  straight  lines  meet  or  intersect  they  are 
said  to  contain,  or  to  make  with  each  other,  an  angle. 

The  two  lines  are  the  sides  of  the  angle,  and  the 
point  of  meeting  is  the  vertex  of  the  angle. 

When  a  line,  coincident  with  one  side  of  an  angle,  re- 
volves about  the  vertex,  always  remaining  in  the  same 
plane,  until  it  arrives  at  the  position  of  the  other  side  of 
the  angle,  the  line  is  said  to  turn. through  the  angle, 
and  the  greater  the  amount  of  turning  the  greater  the 
angle. 

The  magnitude  of  an  angle  may  be  made  clear  by  means  of  a  pair 
of  dividers,  the  legs  of  the  dividers  representing  the  sides  of  the  angle 
and  the  hinge  representing  the  vertex.  If  the  dividers  be  opened  a  given 
amount  an  angle  is  represented;  if  from  that  position  they  be  closed 
to  a  certain  extent  a  smaller  angle  is  represented;  if  they  be  opened 
farther  a  larger  angle  is  represented. 

A  line  may  turn  through  an  angle  in  two  ways,  hence 
there  are  two  angles  which  have  the  same  sides  and  the 
same  vertex. 

For  example,  the  side  O  B  in 
the  figure  may  turn  through  the 
angle  m  by  a  motion  opposite  to 
that  of  the  hands  of  a  watch  to  . 
the  position  O  A,  or  it  may  turn 
through  the  angle  n  by  a  motion 
with  that  of  the  hands  of  a  watch  to  the  same  position  O  A. 

Two  angles  which  have  the  same  vertex  and  the  same 
sides  are  called  conjugate  angles.  When  an  angle  is 
spoken  of,  the  smatterofihe  two  conjugate  angles  is  al- 
ways meant  unless  the  contrary  is  specifically  mentioned. 


DEFINITIONS. 


18.  When  the  two  sides  of  two  conjugate  angles  lie 
in  the  same  straight  line,  the  angle  formed  is  called  a 
straight  angle,  and  the  conjugate  angles  are  equal;  for 

example  the  angle  A  O  B  is  a  straight  angle. 
A o /? 

An  angle  may  be  read  by  reading  the  letter  at  the  ver- 
tex of  the  angle  between  the  letters  upon  the  sides  of 
the* angle,  as  angle  A  B  C  or  angle  CB  A. 
When  there  is  only  one  angle  at  a  given  A 

vertex  it  is  sufficient  to  read  the  letter  at  ^/ 

the  vertex  alone,  as  angle  B.     When  two         ;S 
or  more  angles  have  a  common  vertex  a  ^C^^ 
letter  is  frequently  placed  near  the  vertex         ^*^\^ 
between  the  sides  of  the  angle,  to  desig- 
nate the  angles,  as  m  and  n  in  the  figure 
at   the    right.     For    example,    we    say 
angle    m   instead  of  angle   C  B  A,   and 
angle  n  instead  of  angle  C  B  D. 

19.  Definition.    Two   angles   which 
have  a  common  vertex  and  one  common 
side,  and   are   on  opposite  sides  of  this 
common  side,  are  called  adjacent  angles. 

Angles  m  and  n  in  the  second  figure  in  article  18  are  ad- 
jacent angles. 

20.  Definition.    When   one   straight 
line  meets  another  straight  line  so  that 
the  two  adjacent  angles  formed  are  equal, 
the   angles  are  right  angles.     If  angle 
A  O  B  equals  angle  A  O  C,  the  angles 


A  O  B  and  A  O  C  are  each  right  angles     B        O 


6 


PLANE    GEOMETRY. 


A 


O 


21.  Definition.  An   angle 
less    than    a  right    angle  is 
called  an  acute  angle,  and  an 
angle   greater    than    a   right 
angle    is    called    an    obtuse 
angle.     Angle  B  O  C  is   an 
acute  angle,  and  angle  A  O  B 
is  an  obtuse  angle. 

Acute  and  obtuse  angles  are  called  oblique  angles. 

22.  Definition.  If  a  line  makes  right  angles  with  an- 
other line,  the  two  lines  are  said  to  be  perpendicular  to 
each  other.     In  the  figure  in  article  20,  A  O  is  perpen- 
dicular to  B  C  and  B  C  is  perpendicular  to  A  O. 

23.  Definition.    If  a   line   makes   an   oblique   angle 
with  another  line,  the  two  lines  are  said  to  be  oblique 
to  each  other.     In  the  figure  in  article  21  B  O  is  oblique 
to  A  C  and  A  C  is  oblique  to  B  O. 

24.  When   two   lines   intersect, 
the  opposite  angles  are  called  ver- 
tical angles. 

In  the  figure  at  the  right,  the 
angles  A  O  D  and  COB  are  ver- 
tical angles;  also  the  angles  A  O  C 
and  B  O  D  are  vertical  angles. 

25.  If  two   adjacent  angles  are 
together   equal   to   a   right  angle, 
the  two  angles  are  said  to  be  com- 
plements of  each  other. 

If  the  angles  M  B  D  and  MBA       

are  each  right  angles,  angles  CBD     A  R  D 

and  M  B  C  are  complements  of  each  other;  also  angles 
ABE  and  M  B  E  are  complements  of  each  other. 


DEFINITIONS. 


26.  If  two  adjacent  angles  are  together  equal  to  two 
right  angles,  the  two  angles  are  said  to  be  supplements 
of  each  other. 

In  the  figure  in  article  21,  the  angles  B  O  C  and  A  O  B 
are  supplements  of  each  other. 

Two  adjacent  angles  which  are  supplements  of  each 
other  are  called  supplementary  adjacent  angles. 

LOGICAL  TERMS. 

27.  A  theorem  is  a  truth  which  requires  demonstra- 
tion.    For  example:    If  two  straight  lines  intersect  each 
other  >  the  vertical  angles  are  equal. 

28.  The  statement  of  a  theorem  is  called  its  enuncia- 
tion, or  the  general  enunciation. 

V/hen  a  drawing  is  made  to  illustrate  a  theorem,  the 
description  of  the  drawing  is  called  the  special  enuncia- 
tion. 

29.  A  theorem  consists  of  two  parts,  the  conditions, 
or  suppositions,  and  the  conclusion.     The  conditional 
part  of  a  theorem  is  called  the  hypothesis.     The  hy- 
pothesis is  composed  of  at  least  two  parts:  for  example, 
in  the  theorem  stated  in  article  27,  first>  two  lines  are 
straight,  second,  they  intersect. 

The  two  parts  of  the  hypothesis  are  sometimes  called 
premises. 

30.  The  truth  depending  upon  or  following  from  the 
hypothesis  is  called  the  conclusion. 

In  the  theorem  stated  in  article  27,  the  truth  that  the 
vertical  angles  are  equal  depends  upon  the  hypothesis 
that  two  straight  lines  intersect  each  other. 

31.  The  course  of  reasoning  by  which  the  truth  or 
falsity  of  a  statement  is  established  is  called  the  demon- 
stration, or  proof. 


8  PLANE   GEOMETRY. 

32.  A  problem  is  a  question  proposed  for  solution,  or 
the  statement  of  certain  relations  which  are  to  be  pro- 
duced.    For  instance:    To  construct  a  line  perpendicular  to 
a  given  line  at  a  given  point. 

33.  A  proposition  is  a  general  term  for  a  theorem  01 
a  problem. 

34.  A  corollary  is  a  proposition  easily  deduced  from 
the  proposition  to  which  it  is  attached,  with  the  aid,  if 
necessary,  of  one  or  more  previous  propositions. 

35.  A  scholium  is  a  remark  upon  one  or  more  propo- 
sitions with  respect  to  their  applications,  limitations  or 
connections. 

36.  An  axiom  is  a  truth  which,  from  its  simplicity, 
must  be  admitted  without  demonstration:  as,    The  whole 
of  anything  is  equal  to  the  sum  of  its  parts. 

37.  A  postulate  is  a  problem  whose  solution,  on  ac- 
count of  its  simplicity,  is  admitted  to  be  possible:  as,  Let 
it  be  granted  that  a  straight  line  can  be  drawn  between  two 
points. 

38.  AXIOMS. 

1.  Things  which  are  equal  to  the  same  or  equal  things 
are  equal  to  each  other. 

2.  If  equals  be  added  to  equals  the  sums  will  be  equal. 

3.  If  equals  be  subtracted  from  equals  the  differences 
will  be  equal. 

4.  If  equals  be  multiplied  by  equals  the  products  will 
be  equal. 

5.  If  equals  be  divided  by  equals  the  quotients  will  be 
equal. 


DEFINITIONS.  9 


6.  If  equals  be  added  to  unequals  the  sums  will  be  un- 
equal, and  that  sum  will  be  greater  which  is  obtained 
from  the  greater  magnitude. 

7.  If  equals  be  subtracted  from  unequals  the  differ- 
ences will  be  unequal,  and  that  difference  will  be  greater 
which  is  obtained  from  the  greater  magnitude. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

10.  A  straight  line  is  the  shortest  distance  between 
two- points. 

11.  If  two  straight  lines  coincide  in  two  points  they 
will  coincide  throughout  their  whole  extent,  and  be  one 
and  the  same  straight  line. 

From  this  axiom  it  follows  first,  that  two  straight  lines  can  inter- 
sect in  only  one  point,  and  second,  but  one  straight  line  can  be  drawn 
between  two  points. 

12.  Magnitudes   which   occupy  the   same   space   are 
equal. 

39.  SYMBOLS  AND  ABBREVIATIONS. 

/_    angle.  Ax.  axiom. 

Z^s  angles.  Cons,  construction. 

_L    perpendicular.  Cor.  corollary. 

_Ls  perpendiculars.  Def.  definition. 

||     parallel.  Ex.  exercise. 

||  s  parallels.  Give    auth.  give    authority. 

A    triangle.  Hyp.  hypothesis. 

As  triangles.  Rt.   right. 

£7   parallelogram.  Sch.  scholium. 

£7s  parallelograms.  St.  straight. 

O   circle.  Sug.  suggestion. 

Os  circles.  Sugs.  suggestions. 


10 


PLANE   GEOMETRY. 


PROPOSITION  I.     THEOREM. 

40.  At  a  given  point  in  a  straight  line,  one  perpen- 
dicular can  be  erected  to  that  linef  and  but  one. 


D 


Let  A  B  represent  any  straight  line,  and  O  any  point  in 
the  line. 

I.  To   prove   that   one  perpendicular   can    be   erected 
to  A  B  at  O. 

SUG.  1.  Revolve  a  line,  as  O  C,  from  the  position  O  B 
through  the  st.  Z  A  O  B  to  the  position  O  A.  Art.  17. 

SUG.  2.  In  the  first  part  of  the  revolution,  how 
does  /.COB  compare  with  Z_  A  O  C? 

When  the  revolution  is  nearly  completed,  how  does 
/.COB  compare  with  Z.  A  O  C  ? 

SUG.  3.  Is  there  a  position  of  O  C  during  the  revolu- 
tion at  which  Zs  C  O  B  and  A  O  C  are  equal,  and 
hence  rt.  Zs?  Why? 

II.  To  prove  that  only  one  perpendicular  can  be  erected. 

SUG.  1.  Let  D  O  represent  the  position  of  C  O,  when 
the  two  adjacent  angles  are  equal.  How  much  can  D  O 
be  revolved,  either  way,  and  keep  the  two  adjacent  angles 
equal  ? 


RECTILINEAR    FIGURES. 


11 


SuG.  2.  Then  how  many  _Ls  can  be  drawn  to  the 
line  A  B  at  the  point  O  ?  and  hence  to  any  line  at  any 
point  ? 

Therefore 


41.  COROLLARY  I.     Through  the  ver- 
tex of  a  given  angle  but  one  line  bisect- 
ing the  angle  can  be  drawn. 

SuG.  See  method  in  Art.  40,  I. 

42.  COROLLARY  II.    All  right  angles 
are  equal. 

Let  A  O  C  and  A'  <7  C'  represent  two 
right  angles. 

To  prove  that  they  are  equal. 

SUG.  1.  Place  B  C  upon  B'  C  with 
the  point  O  upon  O '. 

SUG.  2.  Where  will  O  A  fall  with  re- 
spect toVA'l  Why  ?  See  Art.  40,  II. 

Note. — The  pupil  is  expected  to  study  the  sug- 
gestions carefully,  to  follow  the  directions  when 
directions  are  given,  to  answer  the  questions  when 
questions  are  asked,  and  to  give  the  authority  on 


B 


o 


A' 


a      C' 

which  the  answers  are  based;  then  to  go  through  the  whole  demon- 
stration in  a  consecutive  manner,  without  the  aid  of  the  suggestions. 
To  illustrate  what  is  expected  of  the  pupil,  a  model  demonstration 
is  given  of  a  few  propositions,  but  this  model  should  not  be  consulted 
until  after  the  proposition  has  been  studied  by  means  of  the  suggevs 
tions. 


12 


PLANE    GEOMETRY. 


MODEL. 

PROPOSITION  I.    THEOREM. 

43.  At  a  given  point,  in  a  straight  line,  one  perpen- 
dicular can  be  erected  to  the  line,  and  but  one. 


o 


Let  A  B  represent  any  straight  line,  and  O  any  point  in 
the  line. 

I.  To  prove  that  one  perpendicular  can  be  erected 
to  A  B  at  O. 

Suppose  O  C  to  revolve  from  the  position  of  O  B 
through  the  st.  Z.  to  the  position  of  O  A. 

In  the  first  part  of  the  revolution  Z.  B  O  C  is  smaller 
than  Z.  A  O  C. 

When  the  revolution  is  nearly  completed,  as  at  O  C1, 
Z.  B  O  C  is  greater  than  Z  A  O  C' . 

Hence,  as  Z.  B  O  C  is  at  first  small,  and  continually 
increases,  while  Z.  A  O  C  continually  decreases,  there 
must  be  a  position  in  the  revolution  of  O  C  from  O  B 
to  O  A,  as  O  D,  at  which  these  Z$  are  equal,  and  hence 
by  definition  are  rt.  Z&. 

Hence,  by  definition,  O  D  is  _L  to  A  B. 

Therefore,  a  perpendicular  can  be  erected  to  a  given 
line,  at  a  given  point  in  the  line. 


RECTILINEAR    FIGURES. 


13 


II.      To  prove  that  only  one  perpendicular  can  be  erected. 

O  D  is  the  only  perpendicular,  for  if  it  be  revolved 
ever  so  little,  either  way,  one  of  the  angles  becomes 
smaller,  and  the  other  larger,  and  hence  they  are  no 
longer  equal. 

Therefore,  there  can  be  but  one  perpendicular  erected 
to  a  given  line,  at  a  given  point  of  that  line. 

From  parts  I  and  II  of  this  proposition,  it  follows  that: 
At  a  given  point,  in  a  straight  line,  one  perpendicular 
can  be  erected  to  that  line,  and  but  one. 

44.  COROLLARY  I.  Through  the  vertex  of  a  given 
angle,  one,  and  but  one,  straight  line  can  be  drawn  which 
bisects  the  angle. 

Let  the  angle  A  O  C  represent  the 
given  angle. 

If  a  line,  as  O  B,  be  revolved  from  the 
position  O  C,  through  the  given  Z.  to  the 
position  O  A,  there  is  one  position  at 
which  the  adjacent  Z!s  are  equal,  accord- 
ing  to  the  reasoning  in  Prop.  I.  And  if 
from  this  position  O  D  be  revolved  ever  so 
little,  in  either  direction,  one  of  the  Z.§ 
increases,  and  the  other  decreases,  and 
hence  they  can  no  longer  be  equal. 


45.  COROLLARY  II.    All  right  angles    Q 
are  equal. 

Let    the    angles  A  O  B  and  A'  a  ff 
represent  two  right  angles. 

To  prove  that  they  are  equal. 

Place  Z.A  O  C  upon  Z  A'  a  C',  B  C     _ 
upon  B'  C',  and  O  upon  a.  B> 

O  A  must  fall  upon  <7  A'.     Art.  40,  II. 


C' 


14  PLANE    GEOMETRY. 


PROPOSITION  II.     THEOREM. 

46.  If  one  straight  line,  meets  another  straight  line, 
the  sum  of  the  adjacent  angles  formed  equals  two 
right  angles. 


A  B 

Let  C  .B  and  A  J>  represent  two  straight  lines,  wJiich 
meet  at  a  point,  as  B. 

To  prove  that  the  sum  of  the  angles  ABC  and  C  B  D 
equals  two  right  angles. 

SUG.  1.  Suppose,  first,  that  Zs  A  B  Cand  C  B  D  are 
equal.  What  do  you  conclude  concerning  the  number 
of  rt.  Zs  formed?  Why? 

SUG.  2.  Suppose  that  Zs  A  B  C  and  C  B  D  are  un- 
equal, and  let  a  JL,  B  M,  be  erected  to  the  line  A  D,  at 
the  point  B.  Give  authority  for  the  construction. 

SUG.  3.  How  does  the  sum  of  Zs  A  B  C  and  C  B  D 
compare  with  the  sum  of  Zs  ABM  and  M  B  D  ? 

SUG.  4.  What  kind  of  Zs  are  A  B  M  and  MB  Dt 
Why? 

What,  then,  does  the  sum  of  Zs  A  B  C  and  CB  D 
equal  ? 

Therefore 

QUERY.  In  proposition  II  what  is  the  hypothesis? 
What  is  the  conclusion  ? 


RECTILINEAR   FIGURES.  15 

MODEL. 

PROPOSITION  II.     THEOREM. 

47.  If  one  straight  line  meets  another  straight  line, 
the  sum  of  the  two  adjacent  angles  is  equal  to  two 
right  angles. 

M 


A  B 

Let  C  B  meet  A  D,  at  B. 

To  prove  that  the  sum  of  the  angles  A  B  Cand  C  B  D 
equals  two  right  angles. 

If    A  B  C  and    CBD  are    equal,    two   rt.    Zs   are 
formed.  ART.  20. 

If  A  B  C  and  CBD  are  unequal,  draw  B  M  _L  to 
A  D,  at  B.  PROP.  I. 

£ABC  + Z CBD 

=  Z  A  B  M  +  Z  MB  D.     ARTS.  17,  18. 

(Because  each  sum  equals  the  st.  ^  A  B  D.) 

But        Z.A  BM+  Z  MBD  =  Zrt.  Zs.     ART.  22- 
Hence,   Z  A  B  C  +  Z  CB  D  =  2  rt.  Zs.        Ax.  1. 

48.  COROLLARY  I.     The  sum  of  all  the  angles  on  one 
side  of  a  straight  line,  having  a  common  vertex  in  the 
line,  equals  t\w>  right  angles. 

49.  COROLLARY  II.     The   total  angular  magnitude, 
about  a  point,  equals  four  right  angles. 


16  PLANE   GEOMETRY. 

PROPOSITION  III.    THEOREM. 

50.  If  two   straight  lines   intersect,  the   vertical 
angles  formed  are  equal. 


O 

Let  A  B  and  C  Z>  intersect  at  O,  forming  the  vertical 
angles  AO  D  and  COB. 

To  prove  that  angle  A  O  D  equals  angle  COB. 

SUG.  1.  ^AOC+Z.COB=  what  ?     Why? 

Sue.  2.   Z.AOC+Z.AOD  =  what  ?     Why  ? 

SUG.  3.  Compare  Z.  A  O  C  -f  Z.  COB  with  Z.  A  O  C 
+  Z.  A  O  D.     Give  auth. 

SUG.  4.  Compare  Z.  C  O  B  with  Z.  A  O  D.       Ax.  3. 

SUG.  5.  In    a    similar    manner,   compare   ^/s  A  O  C 
and  DOB. 

Therefore 

MODEL. 

PROPOSITION  III.    THEOREM. 

51.  If  two   straight  lines  intersect,  the  vertical 
angles  formed  are  equal. 

A 


Let  A  B  and  C  D  intersect  at  O,  forming  the  vertical 
angles  A  O  D  and  COB. 


RECTILINEAR    FIGURES.  17 

To  prove  that  angle  A  O  D  equals  angle  COB. 

ZAOC+ZCO£=2rt.Zs.  PROP.  II. 
Also  Z  A  O  C  +  Z  A  O  D  =  2  rt.  Zs.  PROP.  II. 
Hence,  /.AOC+Z.COB 

=  ^AOC+Z.AOD.     Ax.  1. 
Hence,  Z.COB=Z.AOD.  Ax.  3. 

In  a  similar  manner  it  may  be  proved  that  4L  A  O  C 
=  /.DOB. 

Therefore,  if  two  straight  lines  intersect,  the  vertical 
angles  formed  are  equal. 

EXERCISES. 

1.  How  many  straight  lines   can   be  drawn  through 
three  points,  not  in  the  same  straight  line,  if  each  line 
connects  two  points  ? 

2.  How  many  straight  lines  can  be  drawn  through  four 
points,  no  three  of  which  are  in  the  same  straight  line, 
if  each  line  connects  two  points  ? 

3.  What  is  the  greatest  number  of  points  in  which 
three  straight  lines  can  intersect  ? 

4.  What  is  the  greatest  number  of  points  in  which 
four  straight  lines  can   intersect  ? 

5.  If  an  angle  is  a  right  angle,  what  is  its  supple- 
ment? 

6.  If  an  angle  is  two  thirds  of  a  right  angle,  what  is 
its  supplement  ? 

7.  If  an  angle  is  three  fourths  of  a  right  angle,  what  is 
its  complement  ? 

8.  In  the  figure  in  Art.  21,  if  the  angle  C  O  B  is  one 
half  of  a  right  angle,  the  angle  A  O  B  equals  what  ? 

9.  In  the  figure  in  Art.  25,  if  the  angle  D  B  C  is  two 
thirds  of  a  right  angle,  and  the  angle  A  B  E  is  three 
fourths  of  a  right  angle,  the  angle  C  B  E  equals  what  ? 

2— Geo. 


18  PLANE  GEOMETRY. 

DEFINITIONS. 

52.  A  plane  triangle  is  a  portion  of  a  plane  bounded 
by  three  straight  lines.     The  points  of  intersection  of  the 
bounding  lines  are  called  the  vertices  of  the  triangle, 
and  the  portions  of  the  bounding  lines  included  between 
the  vertices  are  called  the  sides  of  the  triangle. 

53.  A  right  angled  triangle,  or  simply  a  right  tri- 
angle, is  a  triangle  one  of  whose  angles  is  a  right  angle. 

54.  An  acute  angled  triangle,  or  simply  an  acute 
triangle,  is  a  triangle  all  of  whose  angles  are  acute. 

55.  An  obtuse  angled  triangle,  or  simply  an  obtuse 
triangle,  is  a  triangle  one  of  whose  angles  is  obtuse. 

Obtuse  angled  triangles  and  acute  angled  triangles  are  often  spoken 
of  as  oblique  angled  triangles,  or  simply  oblique  triangles. 

56.  A  triangle,  no  two  sides  of  which  are  equal,  is 
called  a  scalene  triangle. 

57.  A  triangle,  which  has  two  equal  sides,  is  called  an 
isosceles  triangle. 

58.  A  triangle,   which  has   all  three  sides  equal,   is 
called  an  equilateral  triangle. 

59.  The  base  of  a  triangle  is  the  side  upon  which  it  is 
supposed  to  stand.     Generally,  any  side  may  be  assumed 
as  the  base,  but,  in  an  isosceles  triangle,  that  side  which 
is  not  one  of  the  two  equal  sides  is  always  considered 
as  the  base. 

60.  The  angle  which  is  opposite  the  base,  is  called  the 
vertical  angle. 


RECTILINEAR    FIGURES.  19 

PROPOSITION  IV.    THEOREM. 

61.  If  two  triangles  have  two  sides  and  the  in- 
cluded angle  of  one,  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other,  each  to  each,  the  triangles 
are  equal  in  all  respects. 


Let  ABC  and  D  EF  represent  two  triangles,  in  which 
AB=DE,  BC=EF,  and  angle  B  =  angle  E. 

To  prove  that  the  triangles  ABC  and  DBF  are  equal 
in  all  respects,  i.  e. ,  that  A  C  =  D  F,  angle  A  =  angle  D, 
and  angle  C  =  angle  F. 

SUG.  1.  Place  A  A  B  C  upon  A  D  E  F,  so  that  B  C 
coincides  with  E  F,  B  with  E  and  C  with  F. 

SUG.  2.  What  direction  will  B  A  take  with  respect 
to  E  D?  ART.  17. 

SUG.  3.  Where  will  the  point  A  lie  ?     Why  ? 

SUG.  4.  Having  located  the  points  C  and  A,  where, 
with  respect  to  D  F,  does  the  line  A  C  lie  ?  Ax.  11. 

SUG.  5.  What,  now,  is  the  position  of  A  A  B  C  with 
respect  to  A  D  E  Ft 

SUG.  6.  How,  then,  does  A  A  B  C  compare  with 
&DEFt  Ax.  12. 

Therefore 


20  PLANE    GEOMETRY. 

MODEL. 

PROPOSITION  IV.    THEOREM. 

62.  If  two  triangles  have  two  sides  and  the  in- 
cluded angle  of  onet  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other,  each  to  each,  the  triangles 
are  equal  in  all  respects. 

A  P 


Let  ABC  and  &  E  F  represent  two  triangles,  in  which 
=  DE,  BC=EF,  and  angle  B  =  angle  E. 

To  prove  that  the  triangles  ABC  and  D  E  F  are  equal 
in  all  respects,  i.  <?.,  A  C  =  D  F,  Z.  C  =  Z  F,  and 
Z.A  =  Z.D. 

Place  the  A  A  B  C  upon  the  A  D  E  F,  so  that  B  C 
coincides  with  E  F,  B  with  E,  C  with  F. 

Since  Z.  B  =  Z  E,  the  line  B  A  will  take  the  direc- 
tion otE  D.  ART.  17. 

And  since  B  A  =  E  D,  the  point  A  will  fall  upon  D. 

Since  A  falls  upon  D,  and  C  upon  F,  the  side,  A  C, 
will  coincide  with  D  F.  Ax.  11. 

Since  we  have  found  that  the  A  A  B  C  coincides  ex- 
actly with  the  A  D  E  F,  the  two  As  must  be  equal  in 
all  respects.  Ax,  12. 

Therefore,  if  two  triangles  have  two  sides  and  the  in- 
cluded angle  of  one,  equal  to  two  sides  and  the  included 
angle  of  the  other,  the  two  triangles  are  equal  in  all  re- 
spects. 


RECTILINEAR   FIGURES.  21 

PROPOSITION  V.    THEOREM. 

63.  If  two  triangles  liave  two  angles  and  the  in- 
cluded side  of  one,  equal  to  two  angles  and  the  in. 
eluded  side  of  the  other,  each  to  each,  the  triangles 
are  equal  in  all  respects. 


Let  ABC  and  D  E  F  represent  two  triangles,  having 
BC=  EF,  Z.B=  ^E*  and  ^_C=  ^_F. 

To  prove  that  triangles  ABC  and  DBF  are  equal  in 
all  respects. 

SUG.  1.  Place  the  A  A  B  C  upon  the  A  DBF,  so 
that  B  falls  upon  E>  and  C  upon  F.  Why  can  this  be 
done  ? 

SUG.  2.  What  direction  will  B  A  take  ?     Why? 

SUG.  3.  Where  will  A  fall  ?     Why  ? 

SUG.  4.  What  direction  will  C  A  take?     Why? 

SUG.  5.  Where,  now,  will  the  point  A  fall  ?     Why  ? 

SUG.  6.  How,  then,  does  the  A  A  B  C  compare  with 
the  A  D  E  Ft 

Therefore 

SCHOUUM.  When  two  triangles  are  equal,  equal 
sides  lie  opposite  equal  angles,  and  equal  angles  lie  op- 
posite equal  sides. 


22 


PLANE   GEOMETRY. 


PROPOSITION  VI.     THEOREM. 

64.  The  angles  opposite  the  equal  sides  of  an  isos- 
celes triangle  are  equal. 

A 


BMC 

Let  ABC  represent  an  isosceles  triangle,  A  JB  "being 
equal  to  AC. 

To  prove  that  angle  B  is  equal  to  angle  C. 

SUG.  1.  Suppose  A  M  drawn  so  as  to  bisect  Z.  A,  and 
extended  until  it  meets  B  C,  at  M. 
SUG.  2.  Compare  A  A  £Mwiih  A  A  CM.     PROP.  IV. 
SUG.  3.  Compare  Z.  B  with  Z.  C.  Sen.  ART.  63. 

Therefore 


Ex.  10.  If  a  straight  line  bisects  one  of  a  pair  of  vertical 
angles,  prove  that  it  bisects  the  other 
also. 

If  MN  bisects  Z.AOD,  prove  that , ,       \0> 
it    bisects    Z  C  O  B\    that     is,    that' 
Z.COM=Z.BOM.  ^  \£ 

Ex.  11.  If  the  equal   sides  of  an 

isosceles  triangle  be  extended  beyond  the  base,   prove 
that  the  exterior  angles  formed  with  the  base  are  equal. 

Ex.  12.  Points,  in  the  sides  of  an  isosceles  triangle, 
equidistant  from  the  base,  are  equidistant  from  the  vertex. 


RECTILINEAR    FIGURES. 


23 


MODEL. 

PROPOSITION  VI.     THEOREM. 

65.   The  angles,  opposite  the  equal  sides  of  an  isos- 
celes triangle,  are  equal. 


Let  ABC  represent  an  isosceles  triangle,  A  B  being 
eqiuil  to  A  C. 

To  prove  that  angle  B  is  equal  to  angle  C. 

Let  A  M  be  drawn,  bisecting  the  Z.  Ay  and  meeting 
B  C  at  M. 

In  the  &sABM  and  ACM, 

A  B  =  A  C.  HYP. 

Z_BAM=Z_CAM.  CONS. 

And  A  Mis  common. 

Therefore,  the  As  ABM  and  ACM  are  equal  in  all 
respects.  PROP.  IV. 

If  two  ^s  have  two  sides,  and  the  included  ^/  of  one  equal  to  two 
sides,  and  the  included  ^_  of  the  other,  each  to  each,  the  triangles 
are  equal  in  all  respects. 

Therefore,  Z.  B  =  Z.  C,  being  corresponding  ^s  of 
equal  As.  Sen.  ART.  63. 


Ex.  13.  Prove  that  the  line  which  bisects  the  vertical 
angle  of  an  isosceles  triangle,  bisects  the  triangle. 


PLANE    GEOMETRY. 


PROPOSITION  VII.     THEOREM. 

66.  If  a  perpendicular  be  erected  at  the  middle 
point  of  a  straight  line,  the  distances  from  any  point 
in  the  perpendicular  to  the  extremities  of  the  line  are 
equal. 


Let  A  C  represent  any  line,  B  its  middle  point,  B  O,  a 
perpendicular  to  A  C,  at  B,  and  O  any  point  in  the  per- 
pendicular. 

To  prove  that  A  O  and  O  C  are  equal. 

SUG.  1.  In  the  As  O  B  A  and  O  B  C,  what  parts  are 
equal,  each  to  each  ?  Why  ? 

SuG.  2.  How  do  the  As  O  B  A  and  O  B  C  compare  ? 
Why?  ART.  61. 

SUG.  3.  How,  then,  do  O  A  and  O  C  compare  ? 

Therefore 


Ex.  14.  Prove  that  the  bisector  of  the  vertical  angle  of 
an  isosceles  triangle  bisects  the  base,  and  is  perpendicu- 
lar to  the  base. 

Ex.  15.  The  base  of  an  isosceles  triangle,  together  with 
the  bisectors  of  the  angles  at  the  base,  form  a  second 
isosceles  triangle. 


RECTILINEAR    FIGURES.  25 


PROPOSITION  VIII.     THEOREM. 

67.  Any  side  of  a  triangle  is  less  than  the  sum  of 
the  other  two. 


Let  ABC  represent  any  triangle. 

To  prove  that  any  side,  as  A  B,  is  less  than  the  sum  of 
the  other  two. 

SUG.  Which  represents  the  shorter  distance  from  A  to  B, 
that  along  the  line  A  B,  or  along  the  lines  A  C  and  CBt 
Why  ?  Ax.  10. 

Therefore 


Ex.  16.  If  the  angular  magnitude  about  a  point  is 
divided  into  six  equal  angles,  each  angle  is  what  part  of 
a  right  angle  ? 

Ex.  17.  If  the  angular  magnitude  about  a  point  is 
divided  into  three  angles,  the  second  of  which  is  twice 
the  first,  and  the  third  is  three  times  the  first,  how  many 
right  angles  in  each  of  the  three  angles  ? 

Ex.  18.  A  line  which  is  perpendicular  to  the  bisector 
of  an  angle  makes  equal  angles  with  the  sides  of  the 
angle. 


PLANE    GEOMETRY. 


PROPOSITION  IX.     THEOREM. 

68.  If  a  perpendicular  ~be  erected  at  the  middle 
point  of  a  straight  line,  the  distances,  from  a  point 
not  in  the  perpendicular,  to  the  extremities  of  the  line 
are  unequal. 


Let  B  C  represent  a  sfraiyht  line,  M  its  middle  pointf 
O  M,  a  perpendicular  to  1?  C9  at  the  point  M,  A,  a  point  not 
in  the  perpendicular,  and  A  JB  and  A  C  lines  drawn  from 
A  to  the  extremities  of  the  line  It  C. 

7o  prove  that  A  B  and  A  C  are  unequal. 

SuG.  1.  Let  O  be  the  intersection  of  A  B,  and  the  per- 
pendicular. Draw  the  line  O  C. 

SuG.  2.   Compare  A  C  with  A  O  +  O  C.     Give  auth. 
SUG.  3.  Compare  O  C  with  O  B.     Give  auth. 
SuG.  4.  Compare  A  C  with  A  O  +  O  B,  or  with  A  B. 
Therefore 

QUERY.  Why  draw  the  line  O  C? 


Ex.  19.  Prove   that   an    equilateral   triangle   is    also 
equiangular. 

SUG.  See  Prop.  VI. 


RECTILINEAR   FIGURES.  27 

Locus  OF  A  POINT. 

69.  To  be  able  to  fix,  definitely,  the  position  of  a 
point  in  a  plane,  it  is  necessary  to  know  two  things 
about  the  point;  or,  as  it  is  usually  expressed,  two  con- 
ditions limiting  the  position  of  a  point  must  be  given. 
If  only  one  condition  is  given,  the  point  is,  to  some  extent, 
fixed,  but  not  completely  fixed.  For  example,  a  point 
which  is  at  a  given  distance  from  some  fixed  point,  is  it- 
self not  completely  fixed,  but  may  move,  provided  that 
in  its  motion  it  always  satisfies  the  requirement  of  being 
at  the  given  distance  from  the  fixed  point.  The  moving 
point,  then,  may  occupy  any  position  whatever  in  a  line 
which  will  be  defined  later  as  the  circumference  of  a 
circle  whose  center  is  the  fixed  point. 

Definition.  When  the  position  of  a  point  in  a  plane  is 
restricted  in  such  a  way  that  the  point  cannot  be  exactly 
determined,  but  is  limited  to,  and  may  be  anywhere  in, 
a  line,  or  group  of  lines,  the  line,  or  group  of  lines,  is 
called  the  locus  of  the  point. 

The  locus  of  a  point  is  both  inclusive  and  exclusive; 
that  is,  the  line,  or  group  of  lines,  must  include  all  the 
points  which  satisfy  the  given  condition,  and  exclude  all 
points  which  do  not  satisfy  that  condition. 


Ex.  20.  If  two  straight  lines  intersect,  and  one  of  the 
angles  formed  is  a  right  angle,  all  the  angles  are  right 
angles. 

Ex.  21.  If  the  angles  at  the  base  of  an  isosceles  tri- 
angle are  bisected,  the  line  which  joins  the  intersection 
of  the  bisectors  to  the  vertical  angle  bisects  the  vertical 
angle. 


28  PLANE   GEOMETRY. 

PROPOSITION  X.  PROBLEM. 

70,  Determine  the  locus  of  a  point  at  equal  dis- 
tances from  the  extremities  of  a  given  line. 


Let  A  U  represent  the  given  line. 

To  determine  the  locus  of  a  point  at  equal  distances  from 
A  and  B. 

This  problem  means  that  we  are  to  find  the  line,  or 
lines,  which  pass  through  all  those  points  which  are  the 
same  distance  from  A  as  from  B,  and  through  no  other 
points. 

SUG.  1.  What  line  has  all  its  points  equally  distant 
from  A  and  #?  Why? 

SUG.  2.  Are  any  points,  without  said  line,  equally  dis- 
tant from  A  and  B  ?  Why  ? 

SUG.  3.  What,  then,  is  the  locus  of  a  point,  equally 
distant  from  the  extremities  of  A  B,  and  hence  from  the 
extremities  of  any  line  ? 


Ex.  22.  The  line  joining  the  vertices  of  two  isosceles 
triangles,  on  opposite  sides  of  the  same  base,  bisects  the 
base  and  is  perpendicular  to  it. 

Ex.  23.  If  the  straight  line  which  joins  the  vertex  of 
a  triangle  with  the  middle  point  of  the  base  is  perpendic- 
ular to  the  base,  the  triangle  is  isosceles. 

Kx.  24.  Prove  that  a  line  drawn  from  the  vertex  of  an 
isosceles  triangle  to  the  middle  point  of  the  base,  (1)  bi- 
sects the  triangle,  (2)  bisects  the  vertical  angle,  (3)  is 
perpendicular  to  the  base. 


RECTILINEAR   FIGURES. 


29 


PROPOSITION  XI.     THEOREM. 

71.  Two  triangles,  having  the  three  sides  of  one 
equal,  respectively,  to  the  three  sides  of  the  other,  are 
equal  in  all  respects. 

ADA 


A  J$  C  and  D  E  F  represent  iwo  triangles,  having 


To  prore  that  the  triangles  ABC  and  D  E  F  are  equal 
in  all  respects. 

SUG.  1.  Place  the  A  D  E  F  upon  the  A  A  B  C,  so 
that  the  longest  side,  D  F,  of  the  A  D  E  F,  coincides 
with  the  longest  side,  A  C,  of  the  A  A  B  C,  D  upon  A, 
and  F  upon  C,  but  the  point  E  upon  the  opposite  side  of 
A  C,  from  B.  Draw  B  E. 

SUG.  2.  In  the  A  A  B  E,  compare  A  B  with  A  E. 
Give  auth. 

SUG.  3.  Compare  Z.ABE  with  Z.AEB.    Give  auth. 

SUG.  4.  In  the  A  C  B  E,  compare  Z.  C  B  E  with 
Z  C  E  B.  Give  auth. 

SUG.  5.  Compare   Z.  A  B  C   with    Z.  A  E  C. 
auth. 

SUG.  6.  Compare    A  A  B  C   with    A  A  E  C. 
auth. 

SUG.  7.  Then  how  does  'A  A  B  C  compare  with 
A  D  E  Ft  Why  ? 

Therefore  - 


Give 


Give 


30  PLANE    GEOMETRY. 

Ex.  25.  In  Prop.  XI»  place  two  equal  shorter  sides 
upon  each  other,  as  A  B  upon  D  E,  connect  C  and  F, 
and  demonstrate  the  pioposition. 

Ex.  26.  If  two  straight  lines  bisect  each  other  at  right 
angles,  any  point  of  either  is  equidistant  from  the  ex- 
tremities of  the  other. 

Ex.  27.  The  straight  lines  bisecting  the  equal  angles 
of  an  isosceles  triangle  and  terminating  in  the  sides,  are 
equal. 


72.  The  premises  of  a  proposition  are  the  conditions 
given  upon  which  the  conclusion  is  based. 

The  converse  of  a  given  proposition  is  a  proposition 
which  has  the  conclusion  of  the  given  proposition  for  one 
of  its  premises,  and  one  of  the  premises  of  the  given 
proposition  for  its  conclusion. 

Proposition  XII,  separated  into  parts  would  read  as 
follows: 

First  premise  —  There  are  two  adjacent  angles. 

Second  premise  —  Their  sum  equals  two  right  angles. 

Conclusion  —  Their  exterior  sides  form  a  straight  line. 

Proposition  XII  is  the  converse  of  Prop.  II;  which, 
separated  into  its  parts,  would  read  as  follows: 

First  premise  —  There  are  two  adjacent  angles. 

Second  premise  —  Their  exterior  sides  form  a  straight 
line. 

Conclusion  —  Their  sum  equals  two  right  angles. 

When  a  proposition  is  proved  to  be  true,  it  does  not 
necessarily  follow  that  its  converse  is  also  true. 


RECTILINEAR    FIGURES.  31 

PROPOSITION  XII.     THEOREM. 

73.  If  the  sum  of  two  adjacent  angles  equals  two 
right  angles,  their  exterior  sides  form  a  straight  line. 


A~  O  C"  '"M 

Let  OA,  OB,  and  O  C,  be  three  lines,  which  meet  to 
farm,  two  adjacent  angles,  A  O  B  and  B  O  C,  whose  sum 
equals  two  right  angles. 

To  prove  that  the  exterior  sides,  O  A  and  O  C,  form  a 
straight  line. 

SUG.  1.  A  O  C  is  either  a  straight  line  or  a  broken 
line.  To  determine  which  of  these  suppositions  is  true, 
draw  an  extension  of  A  O,  as  O  M. 

SUG.  2.  How  many  rt.  Zs  in  Z  A  O  B  -f  Z  B  O  Ml 
Why? 

SUG.  3.  How  many  rt.  Zs  in  Z.  A  O  B  +  Z  B  O  C  ? 
Why? 

SUG.  4.  Compare  the  sum  of  Z  A  O  B  +  Z.  B  O  M> 
with  the  sum  oiZ.AOB  +  BOC.  Give  auth. 

SUG.  5.  Compare  Z  B  O  C  with  /.BOM.    Give  auth. 

SUG.  6.  Since  the  Zs  B  O  C  and  B  O  M  are  equal, 
and  O  C  and  O  M  are  on  the  same  side  of  O  B,  where 
does  O  C  lie  with  respect  to  O  Ml 

SUG.  7.  But  O  Mis  an  extension  of  A  O,  by  construc- 
tion. What  relation  does  O  C  sustain  to  A  Ol 

Therefore 


32  PLANE    GEOMETRY. 

MODEL. 

PROPOSITION  XII.    THEOREM. 

74.  If  the  sum  of  two  adjacent  angles  equals  two 
right  angles,  their  exterior  sides  form  a  straight  line. 


M 


Let  O  A,  O  B,  and  O  C,  be  three  lines  which  meet  to 
form  two  adjacent  angles,  A  O  B,  and  B  O  C,  whose  sum 
equals  two  right  angles. 

To  prove  that  the  exterior  sides^  O  A  and  O  C,  form  a 
straight  line. 

The  line,  A  O  C,  is  either  a  straight  line  or  a  broken 
line. 

Draw  O  M  an  extension  of  A  O. 

Z  A  O  B  +  Z  B  O  M  =  2  rt.  Zs.         PROP.  II. 
Z  A  O  B  +  Z  B  O  C  =  2  rt.  Zs.  HYP. 

Hence,  /_AOB+^BOM 

=  Z.AOB+Z_BOC.         Ax.  1. 

Hence,  Z.BOM=Z.BOC.  Ax.  3. 

Since  the  Zs  B  O  M  and  B  O  C  are  equal,  and  O  C 
and  (9  M  are  on  the  same  side  of  O  B,  O  C  must  lie  upon 
O  M.  But  A  O  M  is  a  straight  line  by  construction. 
Hence,  as  O  C  lies  upon  O  M,  A  O  C  must  be  a  straight 
line. 

Therefore,  if  the  sum  of  two  adjacent  angles  equals  two 
right  angles,  their  exterior  sides  form  a  straight  line. 


RECTILINEAR    FIGURES.  33 

75.  It  is  often  more  convenient  to  express  the  magni- 
tude of  an  angle  in  some  other  way  than  by  stating  how 
many  right  angles  the  given  angle  contains.  To  obtain 
another  method,  a  right  angle  is  divided  into  ninety 
equal  parts,  called  degrees.  The  magnitude  of  an  angle 
may,  then,  be  expressed  by  stating  how  many  degrees 
the  given  angle  contains. 

EXERCISES. 

28.  How   many   degrees    in    a  straight    angle?     In 
all  the  angular  magnitude  about  a  point  ? 

29.  How   many   degrees   in   the   supplement  of   two 
thirds  of  a  right  angle  ? 

30.  How  many  degrees  in   an   angle  whose  comple- 
ment equals  one  fourth  of  its  supplement  ? 

31.  The    supplement   of  ten   degrees   is   how    much 
more  than  the  complement  of  ten  degrees  ? 

32.  The    supplement    of    any    acute    angle    is    how 
much  more  than  the  complement  of  the  same  acute  angle  ? 

Note. —  In  elementary  geometry  it  is  only  acute  angles  which  have 
complements,  but  either  acute  or  obtuse  angles  have  supplements. 

33.  The  supplement  of  the  complement  of  any  acute 
angle  is  how  much  more  than  the  angle  itself  ? 

34.  The  complement  of  the  supplement  of  an  obtuse 
angle  is  how  much  less  than  the  given  obtuse  angle? 

35.  Two  angles  are  complements  of  each  other,  and 
the  greater  exceeds  the  less  by  38  degrees.     What  are 
the  angles  ? 

36.  If  the  bisectors  of  two  adjacent  angles  are  per- 
pendicular to  each  other,  the  angles  are  supplements  of 
each  other. 

3— Geo. 


34: 


PLANE    GEOMETRY. 


PROPOSITION  XIII.    THEOREM. 

76.  From  a  point  without  a  line,  one,  and  only  one, 
perpendicular  can  be  drawn  to  the  line. 


M 


O 


Let  C  D  represent  the  line,  and  A  the  point. 

I.  To  prove  that  one  perpendicular  can  be  drawn  from 
A  to  the  line  C  D. 

SUG.  1.  Draw  any  line,  as  M  N,  and  at  some  point  of 
this  line,  as  O,  erect  the  JL,  O  P.  PROP.  I. 

SUG.  2.  Place  the  line,  M N,  upon  the  line,  CD,  and 
move  it  back  and  forth  in  C  D. 

SUG.  3.  Is  it  possible  for  O  P  to  embrace  the  point  A  ? 

Therefore 

II-  Only  one  perpendicular  can  lie  drawn  to  the 
line. 


0 


B 

Let  A  O  represent  a  perpendicular  from  the  point,  A,  to 
tJie  line,  C  £>. 


RECTILINEAR    FIGURES.  35 

To  prove  that  no  other  perpendicular  can  be  drawn  from 
the  point  A  to  the  line  C  D. 

SUG.  1.  If  another  _L  can  be  drawn,  let  it  be  repre- 
sented by  A  M. 

SUG.  2.  Extend  A  O  to  D,  making  O  B  =  A  O,  and 
connect  M  and  B. 

SUG.  3.  In  the  As  A  O  M  and  BOM,  compare  A  O 
with^?  O,  M  O  with  M  <9,  and  Z  A  OM\vithZ£  O  M. 

SUG.  4.  Now,  compare  Z  A  M  O  with  Z_  B  M  O. 
Give  auth. 

SUG.  5.  Then,  if  by  construction  Z.  A  M  O  is  a  rt.  Z., 
is  line  A  MB  a  straight  or  a  broken  line?  Why  ? 

SUG.  6.  Then,  how  many  straight  lines  are  drawn 
from  A  to  B  ? 

SUG.  7.  What,  then,  do  you  conclude  about  the  state- 
ment that  A  M  B  is  a  straight  line? 

SUG.  8.  Then,  what  do  you  conclude  as  to  the  possi- 
bility of  A  M  being  a  _L  from  A  to  CD? 

SUG.  9.  Then,  how  many  _Ls  can  be  drawn  from  a 
point  to  a  straight  line  ? 

Therefore 


Ex.  37.  If  a  perpendicular  be  dropped  from  the  vertex 
of  an  isosceles  triangle  to  the  base,  prove,  first,  that  it 
bisects  the  base;  second,  that  it  bisects  the  vertical  angle; 
third,  that  it  bisects  the  triangle. 

Ex.  38.  The  triangles  made  with  the  base  of  an  isos- 
celes triangle,  and  the  lines  bisecting  the  angles  at  the 
base,  are  equal. 

Ex.  39.  A  line  drawn  through  the  vertex  of  an  angle, 
perpendicular  to  the  bisector  of  the  angle,  makes  equal 
angles  with  the  sides  of  the  given  angle. 


PLANE    GEOMETRY. 


MODEL. 

PROPOSITION  XIII  —  PART  II.    THEOREM. 

77.  Only  one  perpendicular  can  be  drawn  from  a 
point  to  a  line. 


M/ 

0 

\ 

X 

H 

Let  A  O  represent  a  perpendicular  from  the  point  A  to 
the  line  C  D. 

To  prove  that  no  other  perpendicular  can  be  drawn  from 
the  point  A  to  the  line  C  D. 

If  another  _L  can  be  drawn,  let  it  be  represented 
by  A  M. 

Extend  A  O  to  D,  so  that  A  O  =  O  B,  and  connect 
Mandfi. 

In  the  As  A  O  M  and  B  O  M,  A  O  =  B  O  (by  cons.), 
MO  =  MO  (identical),  and  ZAOM=  Z  B  O  M 
(each  being  a  rt.  Z_  ).  PROP.  IV. 

If  two  As  have  two  sides  and  the  included  ^/  of  one,  equal  to  two 
sides  and  the  included  ^/  of  the  other,  each  to  each4,  the  triangles  are 
equal  in  all  respects. 

Therefore,  the  As  A  O  M  and  £  O  M  are  equal  in  all 
respects,  and  hence  Z.  A  M  O  =  Z.  D  M  O. 

Since  Z.  A  M  O  is  a  rt.  ^,  by  construction,  and 
Z.  A  M  O  equals  Z.  B  M  O,  A  M  B  is  a  straight 
line.  PROP.  XII. 


RECTILINEAR    FIGURES.  37 

If  two  adjacent  angles  are  together  equal  to  two  right  angles,  their 
exterior  sides  form  a  straight  line. 

But  A  O  B  is  a  straight  line,  hence  A  M  B  cannot  be 
a  straight  line.  Ax.  11. 

Therefore,  only  one  J_  can  be  drawn  from  a  point  to  a 
line. 

Ex.  40.  If  two  vertical  angles   are  bisected   by   two 
straight  lines,  prove   that   the   bi- 
sectors together  form  one  and  the 
same     straight    line.      Prove    that     tf- 
N  O  M  is  a  straight  line. 

SUG.  See  Prop.  XII. 

Ex.  41.  If  A  B  Cbe  an  isosceles  triangle,  whose  ver- 
tex is  A,  and  whose  base  is  B  C,  and  if  ^/be  the  middle 
point  of  A  C  and  N  the  middle  point  of  A  B,  and  if  the 
lines  B  M  and  C  N  intersect  at  O,  prove  (1)  that  B  M 
equals  C  N,  (2)  that  the  triangles  B  C  M  and  C  B  N  art 
equal,  (3)  that  the  triangles  C  O  M  and  B  O  N  are  equal, 
and  (4)  that  the  triangle  B  O  C  is  isosceles. 

Ex.  42.  If  A  B  Cis  an  equilateral  triangle,  and  D,  E, 
and  F  are  points  in  the  sides  A  B,  B  C,  and  C  A,  re- 
spectively, such  that  A  D  =  B  E  =  C  F,  prove  that  the 
triangle  D  E  F  is  equilateral. 

Ex.  43.  If  two  lines,  A  B  and  CD,  intersect  in  the 
point  O,  and  if  the  lines  A  C  and  B  D  be  drawn,  A  B 
-f  C  D  is  greater  than  A  C  +  B  D. 

Ex.  44.  If  A  B  C  and  A  B  D  are  two  triangles  on  the 
same  base,  and  on  the  same  side  of  it,  such  that  A  C 
equals  B  D,  and  A  D  equals  B  C,  and  A  D  and  B  C  in- 
tersect at  6>,  prove  (1)  that  the  triangles  A  B  Cand  A  B  D 
are  equal  in  all  respects,  (2)  that  the  triangles  A  O  C 
and  BOD  are  equal  in  all  respects,  and  (3)  that  the  tri- 
angle A  O  B  is  isosceles. 


38 


PLANE    GEOMETRY. 


78.  Definition.  In  a  right  triangle,  the  side  which  is 
opposite  the  right  angle  is  called  the  hypotenuse. 

PROPOSITION  XIV.    THEOREM. 

79.  Two  rijKt  triangles  which  have  the  hypotenuse 
and  a  side  of  one,  equal  to  the  hypotenuse  and  a  side 
of  the  other,  are  equal  in  all  respects. 


C  DF  ^C          M    B 

Let  ABC  mid  D  E  F  represent  two  right  triangles, 
having  tJie  hypotenuse  A  C  equal  to  the  hypotenuse  D  F9 
and  the  side  A  15  equal  to  the  side  D  E. 

To  prove  that  the  triangles  ABC  and  D  E  F  are  equal 
in  all  respects. 

SuG.  1.  Place  the  triangles  so  that  A  B  coincides  with 
D  E,  A  upon  D,  and  B  upon  E,  with  the  vertices, 
C  and  /%  on  opposite  sides  of  A  B. 

SuG.  2.  Is  C  B  Fa.  straight  or  a  broken  line  ?     Why  ? 

SUG.  3.  Connect  A  with  the  middle  point,  M,  of  C  F. 

SUG.  4.  Compare  As  A  M  C  and  A  M  F.     Give  auth. 

SuG.  5.  Compare  Zs  C  M  A  andFAfA.     Give  auth. 

SuG.  6.  Then,  are  Zs  C  MA  and  F  M  A  right  or 
oblique  Zs  ?  Why  ? 

SUG.  7.  Then,  what  relation  of  position  do  A  M  and 
A  B  sustain  to  each  other  ?  And,  hence,  ^/and  B  ?  Why  ? 

SuG.  8.  Compare  As  ABC  and  A  B  F,  and  hence 
As  A  B  C  and  D  E  F, 

Therefore 


RECTILINEAR    FIGURES. 


39 


PROPOSITION  XV.    THEOREM. 

80.  Two  right  triangles  which  have  the  hypotenuse 
and  an  acute  angle  of  one,  equal  to  the  hypotenuse 
and  an  acute  angle  of  the  other,  are  equal  in  all 
respects. 


Let  ABC  and  D  EF  represent  two  right  triangles 
having  tJie  hypotenuse,  A  B,  equal  to  tlie  hypotenuse,  D  E, 
and  the  angle  A  equal  to  tin  angle  />. 

To  prove  that  the  triangles  ABC  and  D  E  F  are  equal 
in  all  respects. 

SUG.  1.  Place  A  A  B  C  upon  A  D  E  F,  so  that  A  B 
coincides  with  D  E,  A  upon  D,  and  B  upon  E. 

SUG.  2.  What  direction  will  A  C  take  ?     Why  ? 

SUG.  3.  Where  will  the  point  C  fall  ?     Why  ? 

SUG.  4.  Since  B  C  and  E  F  are  both  _L  to  the  line 
D  F,  where  will  B  C  lie  ?  Why  ?  .  PROP.  XIII. 

SUG.  5.  Where  will  the  point  C  fall  ?  (Compare  an- 
swers to  sugs.  3  and  4.) 

SUG.  6.     How,  then,  do  the  two  As  compare  ? 

Therefore 

Note. — In  the  answer  to  Sug.  3  it  will  be  seen  that  C  must  lie 
somewhere  in  the  line  D  Ft  and  in  the  answer  to  Sug.  4  C  must  lie 
somewhere  in  the  line  E  F.  Hence,  in  the  answer  to  Sug.  5,  C  can 
be  exactly  located. 


4:0  PLANE    GEOMETRY. 

81,  Straight  lines,  in  the  same  plane,  that  do  not  and 
cannot  meet,  however  far  extended,  are  called  parallel 
lines. 

PROPOSITION  XVI.     THEOREM. 

82.  Two  lines,  which  are  perpendicular  to  the  same 
line,  are  parallel. 


D 


Let  A  C  and  B  D  represent  two  lines,  each  perpendicu- 
lar to  the  same  line,  A  B,  at  the  points  A  and  B  respec- 
tively. 

To  prove  that  A  C  and  B  D  are  parallel. 

SUG.  1.  The  lines  A  C  and  B  D  will  either  meet  or 
not  meet. 

SUG.  2.  If  they  meet,  represent  the  point  of  meeting 
by  O. 

SUG.  3.  Compare  the  assumption  that  they  meet,  with 
Prop.  XIII. 

SuG.  4.  What,  then,  is  your  conclusion  concerning 
their  meeting  ?  What  do  you  conclude  with  respect  to 
their  being  ||  ?  Give  auth. 

Therefore 


RECTILINEAR    FIGURES.  41 

83-  A  transversal  or  secant  line  is  a  line  which 
crosses  two  or  more  lines;  as  the  line  A  B  in  the  figure. 

When  a  transversal  cuts  two  lines  eight  angles  are 
formed,  viz.;  the  angles  1-8  in  the  figure. 

C Jjb D 


T 


The  interior  angles  are  those  within,  or  between,  the 
lines;  as  3,  4,  5,  6. 

The  exterior  angles  are  those  without  the  lines;  as 
1,  2,  7,  8. 

Alternate  interior  angles  are  pairs  of  interior  angles 
on  opposite  sides  of  the  transversal,  and  not  adjacent;  as 
3  and  6,  5  and  4. 

Alternate  exterior  angles  are  pairs  of  exterior  angles 
on  opposite  sides  of  the  transversal,  and  not  adjacent;  as 
1  and  8,  2  and  7. 

Corresponding  angles  are  pairs  of  angles  on  the  same 
side  of  the  transversal,  one  exterior  and  one  interior  but 
not  adjacent;  as  2  and  6,  4  and  8,  etc. 

Note. — The  two  lints  which  the  transversal  crosses  mayor  may  not 
be  parallel.  If  they  are  not  parallel,  and  are  extended  until  they 
meet,  then  the  two  lines,  together  with  the  transversal,  form  a  tri- 
angle. 


Ex.  45.  The  perpendiculars  from  the  extremities  of 
the  base  of  an  isosceles  triangle  to  the  opposite  sides,  are 
equal, 

Ex.  46.  If  D  &  the  middle  point  of  the  side  B  C,  of  a 
triangle  ABC,  and  B  E  and  C-Fare  the  perpendiculars 
from  B  and  C  to  A  D,  prove  that  B  E  equals  C  F. 


42 


PLANE    GEOMETRY. 


84.  Axiom  13.   Only  one  line  can  be  drawn  through 
a  given  point  parallel  to  a  given  line. 

PROPOSITION  XVII.     THEOREM. 

85.  If  one  of  two  parallel  lines  is  perpendicular  to 
a  given  line,  the  other  one  is  perpendicular  to  the 
same  line. 


Let  E  C  and  F  D  be  two  parallel  lines,  and  let  E  C  be 
perpendicular  to  A  B. 

To  prove  that  F  D  is  perpendicular  to  A  B. 

SUG.  1.  From  some  point  in  F  D,  as  O,  draw  O  M  ±_ 
to  A  B. 

SUG.  2.  What  relation  does  O  M  sustain  to  E  Cl 

PROP.  XVI. 

SUG.  3.  Wh at  relation  does  FD  sustain  to  E  €t    HYP. 

SUG.  4.  What  relation  does  MO  sustain  to  F Dl 

Ax.  13. 

SUG.  5.  Then  what  relation  does  F  D  sustain  to  A  B  ? 
Why?  SeeSug.  1. 

Therefore 


RECTILINEAR    FIGURES.  43 

PROPOSITION  XVIII.     THEOREM. 

86.  If  two  parallel  lines  are  cut  by  a  transversal, 
the  alternate  interior  angles  are  equal. 


U 


Let  MNand  R  P  represent  two  parallel  lines  cut  by  the 
secant  A  B. 

I.  To  prove  that  the  alternate  interior  angles  M  E  B 
and  A  F  P  are  equal. 

SUG.  1.  Through  O,  the  middle  point  of  E  F,  draw 
C  D  _L  to  R  P. 

SUG.  2.  What  relation  does  CD  sustain  to  M  N? 
Why? 

SUG.  3.   Compare  As  O  F  D  and  O  E  C.     Give  auth. 

SUG.  4.  Then,  how  do  Zs  ME  B  and  A  FP  com- 
pare ?  Why  ? 

II.  To  prove  that  the  alternate  interior  angles  NEB 
and  R  F  A  are  equal. 

SUG.  1.  /.MEB+  Z.NEB=  Z.RFA  + ^LPFA. 
Why? 

SUG.  2.  Compare  Zs  NEB  and  R  FA.  (See  Sug. 
4,  Part  I,  and  Sug.  1,  Part  II.) 

Therefore 


44  PLANE   GEOMETRY. 

PROPOSITION  XIX.     THEOREM. 

87.  If  two  parallel  lines  are  cut  by  a  transversal, 
the  corresponding  angles  are  equal. 

A 

c D 


77r 

B 

Let  C  D  and  E  F  represent  tivo  parallel  lines,  cut  by  the 
transversal  A  B. 

To  prove   thai  the  corresponding  angles \   2  -and  6,  or 
3  and  j,  etc. ,  are  equal. 
SUG.  See  Prop.  XVIII. 

PROPOSITION  XX.     THEOREM. 

88.  If  two  parallel  lines  are  cut  l>y  a  transversal, 
the  interior  angles  on  the  same  side  of  the  transversal 
are  supplements  of  each  other. 


-D 


L<et  C  D  and  E  F  represent  two  parallel  lines,  cut  by  the 
transversal  A  B,  and  4  and  6  two  interior  angles  on  the 
same  side  of  A  ~B. 

To  prove  thai  the  angles  4.  and  <5  are  supplement  of  each 
other. 

SUG.  1.  Compare  Zs  3  and  6.     Give  auth. 
SUG.  2.   How  does  Z  3  compare  with  Z  4  ?     Why  ? 
SUG.  3.  How,  then,  does  Z.  6  compare  with  Z.  4  ?  Why  ? 
Therefore 


RECTILINEAR   FIGURES.  45 

PROPOSITION  XXI.    THEOREM. 

89.  If  two  straight  lines  are  cut  by  a  transversal, 
so  that  the  alternate  interior  angles  are  equal,  the 
lines  are  parallel. 


R 


Let  A  B  and  C  D  be  two  straight  lines,  ait  by  the  trans- 
rersal  E F, so  that  the  angles  B  O  F and  ENC are  equal. 

To  prove  that  A  D  and  CD  are  parallel. 

SuG.  1.  Through  O,  the  point  of  intersection  of  A  B 
and  E  F,  draw  R  S  ||  to  C  D. 

SuG.  2.  Compare  Z.  S  O  F  with  Z  E  N  C.  Give 
auth. 

SUG.  3.  Compare  Z.  D  O  F  with  Z.  ENC.  Give 
auth. 

SUG.  4.  Compare  Z.  S  O  F  with  Z.  B  O  F.  Give 
auth. 

SUG.  5.  Since  the  side  O  F  is  the  same  in  the  two 
Z.*  B  O  F  and  S  O  F,  what  relation^pf  position  must 
O  S  and  O  D  sustain  to  each  other? 

SUG.  6.  Since  R  S  is,  by  construction,  ||  to  CD,  what 
relation  does  A  B  sustain  to  C  D  ?  Why  ? 

Therefore 


46  PLANE    GEOMETRY. 

MODEL. 

PROPOSITION  XXI.     THEOREM. 

90.  If  two  straight  lines  are  cut  by  a  transversal, 
&0  that  the  alternate  interior  angles  are  equal,  the 
lines  are  parallel. 

f 

a 


NL 


Let  A  B  and  C  D  be  two  straight  lines,  cut  by  the  trans- 
versal E F,  so  tJuit  the  angles  B  O  F  and  ENC are  equal. 

To  prove  that  A  B  and  C  D  are  parallel. 

Through  O,  the  point  of  intersection  of  A  B  and  E  Fy 
drow  R  S  ||  to  C  D. 

The  Z.  S  O  F  equals  Z  E  N  C. 

(If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate  interior 
angles  are  equal.) 

But  Z  B  O  F  equals  Z  E  N  C.     By  Hyp. 

Hence,  Z.  S  O  /^equals  Z.  B  O  F. 

(Things  equal  to  the  same  thing  are  equal  to  each  other.) 

Since  the  side  O  F  is  the  same  in  the  equal  Z§  S  O  F 
and  B  O  F,  OS  and  O  B  must  coincide. 

But  R  S  is,  by^ons.,  ||  to  CD.  Hence  A  B,  which 
coincides  with  R  S,  must  be  ||  to  CD. 

Therefore,  if  two  straight  lines  are  cut  by  a  transversal, 
so  that  the  alternate  interior  angles  are  equal,  the  lines 
are  parallel. 


RECTILINEAR   FIGURES.  47 

PROPOSITION  XXII.    THEOREM. 

91.  If  iwo  straight  lines  are  cut  by  a  transversal 
so  that  the  corresponding  angles  are  equal,  the  lines 
are  parallel. 


-B 


c Of. o 


Let  A  B  and  C  J>  represent  two  lines  cut  by  the  trans- 
versal E F,  so  that  the  corresponding  angles  EO  A  and 
E  N  C  are  equal. 

To  prove  that  A  B  and  C  D  are  parallel. 

SuG.  1.  Through  O,  the  point  of  intersection  of  A  B 
and  E  F,  draw  R  S  to  represent  a  line  ||  to  C  D. 

SuG.  2.  Compare  Z.  E  O  R  with  Z.  E  N  C.  Give 
auth. 

SUG.  3.  Compare  Z.  E  O  A  with  Z.  E  N  C.   Give  auth. 

SUG.  4.   Compare  Z.  E  O  R  with  Z.  E  O  A.   Give  auth. 

SUG.  5.  Since  the  side  E  O  is  the  same  in  the  two 
Z.s  E  O  A  and  E  O  R,  what  relation  of  position  must 
O  A  and  O  R  sustain  to  each  other  ? 

SUG.  6.  Since  R  S  is,  by  construction,  ||  to  CD,  what  re- 
lation does  A  B  sustain  to  C  D  ?  Why  ? 

Therefore 


4:8  PLANE    GEOMETRY. 

PROPOSITION  XXIII.     THEOREM. 

92.  //  two  straight  lines  are  cut  by  a  transversal  so 
that  the  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplements  of  each  othert  the  lines  are 
parallel. 


,B 

/           -s 

£ 

OL.  g 

R  

/ 

/ 

c          OL 

D 

Let  A  B  and  C  D  represent  two  lines  cut  l>y  the  trans- 
versal E  F9  so  that  the  angles  A  O  F  and  E  J\  C  are  sup- 
plements of  each  other. 

To  prove  ihat  A  B  and  C  D  are  parallel. 

SUG.  1.  Through  O,  the  point  of  intersection  of  A  D 
and  E  F,  draw  R  S  to  represent  a  line  ||  to  C  D. 

Suo.2.  Compare  Z.AOP  with  Z.ROF.  (See  method 
in  Prop.  XXI.) 

Complete  the  demonstration. 

There  lore 

93.  An  exterior  angle  of  a 
triangle  is  an  angle  formed  by  one 
side  of  a  triangle  and  an  adjacent 
side  extended.  Angle  n  is  an  ex- 
terior angle  of  the  triangle  ABC. 


RECTILINEAR    FIGURES.  49 

94.  When  one  side  of  a  triangle  has  been  extended  to 
form  an  exterior  angle,  the  two  interior  angles  of  the  tri- 
angle which  are  not  adjacent  to  the  exterior  angle  are 
called  opposite  interior  angles.     In  the  figure  in  the 
previous  article  n  is  an  exterior  angle,  and  the  angles 
A  and  C  are  the   opposite   interior  angles  of  the  tri- 
angle ABC. 

PROPOSITION  XXIV.     THEOREM. 

95.  An  exterior  angle  of  a  triangle  equals  the  sum 
of  the  opposite  interior  angles. 


BC  represent  a  triangle,  D  AC  an  exterior  anglet 
and  B  and  C  the  opposite  interior  angles. 

To  prove  that  the  angle  D  A  C  equals  the  sum  of  the 
angles  B  and  C. 

SUG.  1 .  Through  the  point  A  draw  a  line  M  N  ||  to  B  C. 
SUG.  2.   Compare  /.DAN  with  Z.  B.     Give  auth. 
SUG.  3.  Compare  Z.  C  A  A7' with  Z.  C.     Give  auth. 
SUG  4.  Compare  Z  D  A  C  with  Z  B  -f  Z  C. 
Therefore 

4— Geo. 


50  PLANE    GEOMETRY. 


PROPOSITION  XXV.     THEOREM. 
96.  The  sum  of  the  interior  angles  of  a  triangle 
equals  two  right  angles. 


'~       ~"D  B 

Let  ABD  represent  a  triangle. 

To  prove  that  the  sum  of  the  angles  A,  B  and  B  D  A 
equals  two  right  angles. 

SuG.  1.  Extend  one  of  the  sides,  as  B  D. 
SuG.  2.  Apply  Prop.  XXIV  and  complete  the  demon- 
stration. 
Therefore 

97.  COROLLARY  I.     A  triangle  cannot  have  more  than 
one  obtuse  angle. 

98.  COROLLARY  II.    Every  right  triangle  has  two  acute 
angles,  each  of  which  is  the  complement  of  the  other. 

PROPOSITION  XXVI.    THEOREM. 

99.  If  two  triangles  have  two  angles  of  one  equal 
respectively  to  two  angles  of  the  other,  the  third  angles 
are  equal. 

SUG.  Apply  Prop.  XXV  to  each  A  and  complete  the 
demonstration. 

100.  COROLLARY.     If  two  right  triangles  have  an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other,  the  re- 
maining acute  angles  are  equal. 


RECTILINEAR   FIGURES. 


51 


PROPOSITION  XXVII.    THEOREM. 

101.  If  two  right  triangles  have  a  side  and  an  acute 
angle  of  one  equal  to  a  side  and  an  acute  angle  of  the 
other,  the  triangles  are  equal  in  all  respects. 


"Let  ABC  and  D  E  F  represent  two  right  triangles  in 
which  A  C  equals  D  F,  and  ^_  B  equals  ^_  E. 

To  prove  that  the  triangles  ABC  and  DBF  are  equal 
in  all  respects. 

Sue.  Compare  Zs  A  and  D.     Give  auth. 
Complete  the  demonstration. 
Therefore 


Ex.  47.  Prove  Prop.  XXVII  by  taking  the  side  B  C 
and  the  angle  A,  equal  to  the  side  E  F  and  the  angle  D 
respectively;  also,  by  taking  the  side  A  B  and  the  angle 
B,  equal  to  the  side  D  E  and  the  angle  E  respectively. 

PROPOSITION  XXVIII.    THEOREM. 

102.  If  two  triangles  have  two  angles  and  a  side  of 
one  equal  respectively  to  two  angles  and  a  correspond- 
ing side  of  the  other f  the  triangles  are  equal  in  all  re- 
spects. 

Draw  a  figure  and  give  the  demonstration. 


52  PLANE   GEOMETRY. 


PROPOSITION  XXIX.     THEOREM. 

103.  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  them  cure  equal  and  the  triangle  is  isosceles. 


B  M  C 

Let  ABC  represent  a  triangle  in  which  the  angle  B 
equals  the  angle  C. 

To  prove  that  the  side  A  C  equals  the  side  A  B. 

SUG.  1.  Draw  a  line  A  M  to  represent  a  J_  from  A 
io£C 

SUG.  2.  Compare  As^4  MB  and  A  M  C.     Give  auth. 
SUG.  3.  Compare  A  C  with  A  B. 
Therefore 


Ex.  48.  If  two  parallel  lines  are  cut  by  a  transversal, 
prove  that  the  alternate  exterior  angles  are  equal. 

Ex.  49.  If  the  middle  points  of  the  three  sides  of  an 
isosceles  triangle  be  taken  for  the  three  vertices  of  a  sec- 
ond triangle,  prove  that  the  second  triangle  is  isosceles. 

Ex.  50.  If  from  any  point  in  the  base  of  an  isosceles 
triangle  perpendiculars  be  drawn  to  the  sides,  these  per- 
pendiculars make  equal  angles  with  the  base. 

Ex.  51.  In  the  figure  for  Prop.  XX,  compare  angles 
1  and  6,  also  angles  4  and  7. 


RECTILINEAR   FIGURES.  53 

PROPOSITION  XXX.    THEOREM. 

104.  //  two  angles  of  a  triangle  are  unequal,  the 
sides  opposite  them  are  unequal,  the  greater  side  being 
opposite  the  greater  angle. 


Let  ABC  represent  a  triangle  in  which  the  angle  C  is 
greater  than  the  angle  B. 

To  prove  that  the  side  B  A  is  greater  than  the  side  C  A. 

SUG.  1.  Draw  C  M  to   represent   a  line   making   the 
Z  B  CM  equal  to  the  Z  B. 
SUG.  2.  Compare  B  Mand  CM.     Give  auth. 
SUG.  3.  Compare  CM  -f  MA  with  C  A.     Give  auth. 
SUG.  4.  Compare  B  M  +  MA  with  C  A. 
SUG.  5.  Compare  B  A  with  C  A. 
Therefore 


Ex.  52.   Prove  Prop.  XXV  by  another  method. 

SUG.  Through  A  draw  a  line  ||  to  D  B. 

Ex.  53.  In  Prop.  XXV  extend  D  A  and  B  A  through 
A,  and  draw  through  A,  a  line  ||  to  O  B,  and  then  prove 
the  proposition  in  another  way. 


54  PLANE   GEOMETRY. 

PROPOSITION  XXXI.    THEOREM. 

105.  If  two  sides  of  a  triangle  are  unequal  the 
angles  opposite  them  are  unequal,  the  greater  angle 
being  opposite  the  greater  side. 

A 


jy  c 

Let  ABC  represent  a  triangle  in  which  the  side  A  Bis 
greater  than  the  side  A  C. 

To  prove  that  the  angle  C  is  greater  than  the  angle  B. 

SUG.  1.  The  Z  C  equals  Z  B,  or  is  less  than  Z  B,  or 
is  greater  than  Z  B. 

SUG.  2.  If  Z  C  were  equal  to  Z  B  how  would  A  B 
and  A  C  compare  ?  Why  ?  Does  this  agree  with  the  hyp .  ? 
Is  it  possible  for  Z*  C  to  equal  Z  B  ? 

SUG.  3.  If  Z  C  were  less  than  Z  B  how  would  A  B 
and  A  C  compare  ?  Why  ?  Does  this  agree  with  the  hyp.  ? 
Is  it  possible  for  Z  C  to  be  less  than  Z  B  ? 

SUG.  4.  From  the  results  reached  in  Sugs.  1,2,3,  what 
must  be  the  relation  between  Zs  C  and  B  ? 

Therefore  - 

Note. — If  the  student,  in  original  investigation  of  the  proposition 
just  given,  or  any  similar  one,  should  chance  to  consider  the  supposi- 
tion which  leads  to  the  truth  before  one  or  more  of  the  others,  the  re- 
maining suppositions  should  be  investigated.  When  all  possible  sup- 
positions have  been  stated,  one  of  which  is  true,  and  it  has  been  shown 
that  all  but  one  are  false,  it  is  evident  that  the  one  remaining  must  be 
true. 


RECTILINEAR   FIGURES.  55 

106.  Proposition  XXXI  is  a  good  illustration  of  what  is 
known  as  the  indirect  method,  or  the  reductio  adabsurdum 
method  of  reasoning.  Its  peculiarity  consists  in  the  fact 
that  the  statement  of  the  proposition  is  not  directly 
proved  to  be  true,  but  that  everything  which  contradicts 
the  statement  of  the  proposition  is  shown  to  lead  to  some 
manifest  absurdity,  and  is  therefore  false.  This  method 
often  presents  difficulties  to  the  beginner  on  account  of 
the  fact  that  he  is  obliged  to  admit,  temporarily,  and  for 
argument's  sake,  something  which  the  argument  itself 
goes  to  destroy.  But,  as  the  reasoning  is  examined  more 
closely,  it  will  be  found  to  be  just  as  satisfactory  as  the 
direct  demonstration,  and  in  many  cases  (especially  the 
converse  of  propositions  which  have  been  proved  by  di- 
rect demonstration),  this  method  is  the  easiest  and  best 
one  to  apply.  In  applying  this  method  care  must  be 
taken  that  every  possible  case  which  contradicts  the  propo- 
sition be  considered  and  each  one  shown  to  lead  to  an  ab- 
surdity. Then,  and  then  only,  is  this  method  of  demon- 
stration rigid. 


Ex.  54.  If  A  B  C  be  a  right  triangle  with  the  right 
angle  at  C,  and  if  through  C  a  line  meeting  the  hypote- 
nuse at  D  be  drawn  in  such  a  manner  that  the  angle 
A  C  D  equals  the  angle  B  and  the  angle  BCD  equals 
the  angle  A,  prove  that  CD  is  perpendicular  to  the 
hypotenuse  A  B. 

Ex.  55.  If  A  B  C  be  a  right  triangle  with  the  right 
angle  at  C,  and  if  through  C  a  line  meeting  the  hypote- 
nuse at  D  be  drawn  in  such  a  manner  that  the  angle  A  CD 
equals  the  angle  A,  and  the  angle  BCD  equals  the 
angle  B,  prove  that  CD  bisects  the  hypotenuse  A  B. 


56 


PLANE   GEOMETRY. 


PROPOSITION  XXXII.    THEOREM. 

107.  If  a  perpendicular  and  oblique  lines  are 
drawn  from  a  point  to  a  given  line: 

I-  The  perpendicular  is  shorter  than  any  oblique 
line. 

II-  Two  oblique  lines  which  meet  the  given  line  at 
equal  distances  from  the  foot  of  the  perpendicular, 
are  equal. 

III.  Of  two  oblique  lines  meeting  the  given  line  at 
unequal  distances  from  the  foot  of  the  perpendicular, 
the  more  remote  is  the  greater. 


Let  AD  be  perpendicular  to  a  given  line  J5  E9  and  A  &, 
A  C  and  A  E  oblique  lines,  meeting  the  given  line  at  J5,  C 
and  E  respectively;  and  let  DBbe  greater  tJian  D  E,  and 
Z>  C  equal  to  D  E. 

I.  To  prove  that  A  D  is  the  shortest  line  from  A  to  B  E. 

SuG.  1.  If  A  C  represents  any  oblique  line,  how  does 
Z.  A  D  C  compare  with  Z  A  C  D*  Why? 

SUG.  2.  How  does  A  D  compare  with  A  C,  and  hence 
with  any  oblique  line  from  A  to  B  EJ 

II.  To  prove  that  A  C  equals  A  E. 

SuG.   Compare  the  triangle  ADC  and  A  D  E. 


RECTILINEAR    FIGURES.  57 

Complete  the  demonstration  of  this  part  of  the  propo- 
sition. 

III.      To  prove  that  A  B  is  greater  than  A  E. 

SUG.  1.  Compare  ^.s  A  B  D  and  A  CD  with  respect 
to  magnitude.     Give  auth. 

SUG.  2.  Compare  Zs  A  CD  and  A  C B  with  respect 
to  magnitude.     Give  auth. 

SUG.  3.  Compare  Zs  A  C  B  and  ABC  with  respect 
to  magnitude. 

SUG.  4.  Compare  A  B  with  A  C,  and  hence  with  A  E. 
Give  auth. 

SUG.  5.  Notice  the  conclusions  in  parts  I,  II  and  III. 

Therefore 


Ex.  56.  A  line  drawn  from  one  end  of  the  base  of  an 
isosceles  triangle,  perpendicular  to  the  opposite  side, 
makes  with  the  base,  an  angle  equal  to  one  half  the  ver- 
tical angle. 

Ex.  57.  A  straight  line  drawn  from  any  point  in  the 
bisector  of  an  angle  to  either  side  and  parallel  to  the 
other  side,  makes,  with  the  bisector  and  the  side  to  which 
the  line  is  drawn,  an  isosceles  triangle. 

Ex.  58.  The  angle  formed  by  the  bisectors  of  the 
angles  at  the  base  of  an  isosceles  triangle  is  equal  to  an 
exterior  angle  at  the  base  of  the  triangle. 

Ex.  59.  A  C  B  and  A  D  B  are  two  triangles  on  the 
same  side  of  A  B,  such  that  A  C  equals  B  D,  and  A  D 
equals  B  C,  and  A  D  and  B  C  intersect  at  O.  Prove 
that  A  O  B  is  an  isosceles  triangle. 

Ex.  60.  The  difference  between  any  two  sides  of  a  tri- 
angle is  less  than  the  third  side. 


58  PLANE   GEOMETRY. 

PROPOSITION  XXXIII.    THEOREM. 

108.  Converse  of  Prop.XXXII,!.  The  shortest  line 
from  a  point  to  a  line  is  a  perpendicular  to  the  line. 


M        B 

"Let  A  B  represent  the  shortest  line  from  A  to  the  line  C  D. 

To  prove  that  A  B  is  perpendicular  to  C  D. 

SUG.  1.  If  A  B  is  not  _L  to  C  D,  draw  A  M  to  repre- 
sent a  _L  to  C  D. 

SUG.  2.  If  A  M  is  J_  to  C  D,  how  does  it  compare  in 
length  with  A  B1  Prop.  XXXII. 

SUG.  3.  Compare  your  answer  with  the  hypothesis. 

Therefore 

109.  The  distance  from  a  point  to  a  line  is  the 
length  of  the  perpendicular  from  the  point  to  the  line. 


Ex.  61.  Lines  which  are  perpendicular  to  parallel  lines 
are  parallel. 

Ex.  62.  If  the  equal  angles  of  an  isosceles  triangle  be 
bisected,  the  bisectors  and  the  base  of  the  triangle  form 
a  new  isosceles  triangle  whose  vertical  angle  is  equal  to 
an  exterior  angle  at  the  base  of  the  original  triangle. 

Ex.  63.  The  sum  of  the  three  sides  of  any  triangle  is 
greater  than  the  double  of  any  one  side,  but  less  than 
the  sum  of  the  doubles  of  any  two  sides. 


RECTILINEAR   FIGURES. 


59 


PROPOSITION  XXXIV.    THEOREM. 
110.  Two  equal  oblique  lines,  drawn  from  the  same 
point  in  the  perpendicular  to  a  given  linef  cut  off 
equal  distances  from  the  foot  of  the  perpendicular. 


C 


B 


D 


Let  ABbe  perpendicular  to  C  D,  and  let  AC  and  A 
be  equal  oblique  lines  drawn  from  A  to  CD. 

To  prove  that  B  C  equals  B  D. 

SUG.  Compare  As  A  B  C  and  A  B  D.     Give  auth. 
Therefore 


Ex.  64.  Each  angle  of  an  equilateral  triangle  is  one 
third  of  two  right  angles,  or  two  thirds  of  one  right 
angle. 

Ex.  65.  The  line  joining  the  feet  of  the  perpendiculars 
drawn  from  the  extremities  of  the  base  of  an  isosceles 
triangle  to  the  opposite  sides  is  parallel  to  the  base. 

Ex.  66.  The  sum  of  the  three  straight  lines  drawn  from 
any  point  within  a  triangle  to  the  three  vertices,  is  less 
than  the  sum  and  greater  than  the  half  sum  of  the  sides. 

Ex.  67.  If  a  line  intersects  the  sides  of  an  isosceles 
triangle  at  equal  distances  from  the  vertex,  the  line  is 
parallel  to  the  base. 


60 


PLANE    GEOMETRY. 


PROPOSITION  XXXV.     THEOREM. 

111.  Two  unequal  oblique  lines,  drawn  from  the 
same  point  in  the  perpendicular  to  a  given  line,  cut 
off  unequal  distances  from  the  foot  of  the  perpendicu- 
lar, the  longer  line  cutting  off  the  greater  distance. 


Let  ABbe  perpendicular  to  C  D,  and  let  A  C  and  A  D 
be  unequal  oblique  lines  drawn  from  A  to  C  D,  A  C  being 
longer  than  A  D. 

To  prove  that  B  C  is  greater  than  B  D. 

SUG.  1.  B  Cis  equal  to  B  D,  or  is  less  than  B  D,  or  is 
greater  than  B  D. 

SUG.  2.  If  B  C  were  equal  to  B  D,  how  would  A  C 
compare  with  A  D  ?  Why  ? 

SUG.  3.  If  B  C  were  less  than  B  D,  how  would  A  C 
compare  with  A  D  ?  Give  auth. 

SUG.  4.  How,  then,  must  B  C  compare  with  B  Dl 

Therefore 


Ex.  68»  If  the  vertical  angle  of  an  isosceles  triangle  is 
one  half  as  great  as  an  angle  at  the  base,  a  bisector  of'  a 
base  angle  divides  the  given  triangle  into  two  isosceles 
triangles. 


RECTILINEAR  FIGURES.  61 

EXERCISES. 

69.  Prove   that   every   point   in   the    bisector 
angle  is  equally  distant  from  the  sides  of 

the  angle. 

SUG.  O  M  and  O  N  are  J_  to  A  B  and 
A  C  respectively.  (See  definition  of  dis- 
tance.) 

Prove  O  M  equal  to  O  N. 

70.  Prove  that  every  point  not  in  the 
bisector  of  an  angle  is  unequally  distant 
from  the  sides  of  the  angle. 

Prove  that  O  M  and  O  N  are  unequal. 

SUG.  Draw  S  R  _L  to  A  N.  Connect 
R  and  O. 

Compare  5  O  +  S  R  with  OR,  O  M 
with  OR,  O  M  with  O  N. 

71.  What  is  the  locus  of  a  point  equally  distant  from 
the  sides  of  an  angle  ? 

72.  What  is  the  locus  of  a  point  at  equal  distances 
from  the  two  intersecting  lines  ? 

73.  The  middle  point  of  the  hypotenuse  of  a  right  tri- 
angle is  at  equal  distances  from  the  three  vertices. 

74.  The  sum  of  the  exterior  angles  at  the  base  of  any 
triangle   is  equal  to  two  right  angles  plus  the  vertical 
angle  of  the  triangle. 

75.  In  a  triangle  ABC,  the  angle  C  is  twice  the  sum 
of  the  angles  A  and  B,  and  the  angle  B  is  twice  the 
angle  A\  find  all  three  angles  of  the  triangle. 

76.  In  a  triangle  ABC,  the  angle  B  is  three  times  the 
angle  A,  and  the  angle  C  is  five  times  the  angle  A\  find 
each  angle  of  the  triangle. 


62  PLANE   GEOMETRY. 

PROPOSITION  XXXVI.     THEOREM. 

112.  If  two  lines  be  drawn  from  a  point  in  a  tri- 
angle to  the  extremities  of  a  side,  the  sum  of  the  two 
lines  so  drawn  is  less  than  the  sum  of  the  ottier  two 
sides  of  the  triangle. 


Let  ABC  represent  any  triangle,  O  any  point  within 
the  triangle,  and  O  B  and  O  C  lines  drawn  from  O  to  B 
and  C,  respectively. 

To  prove  that  O  B  +  O  Cis  less  than  A  B  +  A  C. 

SUG.  1.  Extend  one  of  the  lines  through  O  to  meet  the 
side  of  the  triangle  at  M. 

SUG.  2.  How  does  B  A  +  A  M  compare  with  B  Ml 
Why? 

SUG.  3.  Then  how  does  BA  +  AM+MC  compare 
with^^/+  M  C? 

SUG.  4.  Compare  O  M  +  M  C  with  O  C. 

SUG.  5.  Then  how  does  B  O  +  O  M  +  M  C  compare 
with  B  O  +  O  C? 

SUG.  By  referring  to  your  answers  to  Sugs.  3  and  5, 
compare  B  A  +  A  C  with  B  O  +  O  C. 

Therefore 


RECTILINEAR  FIGURES.  63 

PARTIAL  DEMONSTRATION. 

Answer  to  Sug.  3.  B  A  +  A  C  is  greater  than  B  M 
+  MC. 

(If  equals  be  added  to  unequals  the  results  are  unequal,  that  being 
greater  which  is  obtained  by  addition  to  the  greater  of  the  two  given 
unequals.) 

Answer  to  Sug.  5.  B  M  +  M  C  is  greater  than  B  O 
•f  O  C.  (Same  reason.) 

Answer  to  Sug.  6.  Hence,  much  more  must  BA 
+  A  C  be  greater  than  B  O  +  O  C. 


Ex.  77.  Prove  that  the  three  bisectors  of  the  angles  of 
a  triangle  meet  in  the  same  point. 

SUG.  1.  Draw  a  A  A  B  C. 

SUG.  2.  The  bisector  of  the  Z  A  is  the  locus  of  a 
point  equally  distant  from  the  sides  A  B  and  A  C.  See 
exercises  69  and  70. 

SUG.  3.  The  bisector  of  the  Z  B  is  the  locus  of  what? 

SUG.  4.  Prove,  now,  that  the  point  of  intersection  of 
these  two  bisectors  is  on  the  bisector  of  the  /_  C. 

Ex.  78.  If  a  straight  line  intercepted  between  parallel 
lines  is  bisected,  any  other  straight  line  drawn  through 
the  point  of  bisection  to  meet  the  parallel  lines  is  also  bi- 
sected at  that  point;  and  the  two  intersecting  lines  cut 
off  equal  portions  on  the  parallel  lines. 

Ex.  79.  A  B  is  the  hypotenuse  of  a  right  triangle  ABC, 
B  D  is  drawn  bisecting  the  angle  B,  meeting  A  C  at  D, 
and  D  E  is  drawn  perpendicular  to  A  C,  meeting  A  B 
at  E.  Prove  that  E  D  B  is  an  isosceles  triangle. 


64  PLANE   GEOMETRY. 

PROPOSITION  XXXVII.    THEOREM. 

113.  If  two  triangles  have  two  sides  of  one  equal  to 
two  sides  of  the  other  and  the  included  angles  un- 
equal, the  remaining  sides  are  unequal,  that  "being 
greater  which  belongs  to  the  triangle  having  the 
greater  included  angle. 


Let  ABC  and  A'  B'  €'  represent  two  triangles,  in 
which  A  B  equals  A'  B',  A  C  equals  A'  C't  and  ^_  A  is 
greater  tfian  ^_A'. 

To  prove  that  B  C  is  greater  than  B'  C'. 

SUG.  1.  Place  A  A'  B1  C'  upon  A  A  B  C  so  that  A'  & 
coincides  with  A  B,  A'  upon  A,  and  B1  upon  B. 

SUG.  2.  Since  ^  A'  is  less  than  ^  A,  where,  with  re- 
spect to  Z  B  A  C,  must  A'  C'  fall  ? 

CASE  I.    Suppose  that  C  falls  without  the  triangle  ABC. 

SUG.  3.  Bisect  the  Z.  C  A  C',  and  extend  the  bisector 
to  meet  B  C  at  O.  Connect  O  and  C'. 

SUG.  4.  Compare  As  A  O  C  and  A  O  C'.    Give  auth. 

SUG.  5.  Compare  O  C  and  O  C. 

SUG.  6.  Compare  B  O  +  O  C'  with  B  C,  hence  B  C 
with  B  C',  hence  B  C  with  B1  C'. 

SUG.  7.  What  is  your  conclusion  in  this  case  ? 


RECTILINEAR   FIGURES. 


65 


CASE  II.     Suppose  that  C'  falls  upon  the  line  B  C. 

SUG.  8.  In  the  figure  at  the  A 

right  compare  B  C'  with  B  Ct 
hence  B'  C'  with  B  C. 

(A  part  is  less  than  the  whole.) 

SUG.  9.  What  is  your  con- 
clusion in  this  case  ?  B  C'  ~"^ 

CASE  III.  Suppose  that  Cf falls  within  the  triangle  ABC. 

SUG.  10.  In  the  figure 
at  the  right  compare  A  C 
+  C^with  A  C+C  B. 

SUG.  11.  As^Cequals 
A  C',  how  does  B  C' 
compare  with  B  C,  and 
hence  B1  C'  with  B  C? 

SUG.  12.  What  is  your 
conclusion  in  this  case? 

Therefore 

Note. — The  pupil  can  read- 
ily see  the  possibility  of  the 
three  cases  in  this  proposition 
by  an  examination  of  the  fig- 
ure at  the  right. 


B 


Ex.  80.  Give  another 
demonstration  of  Case  I,  of  Proposition 
XXXVII,  by  using  the  following  sug- 
gestions: 

SUG.  1.  Connect  Cand  C'.  B 

SUG.  2.  Compare  Z.  A  C  Cwith  A  CC. 

SUG.  3.  Compare  ^LBC  Cwith  B  CC. 

Therefore 

5— Geo. 


66  PLANE    GEOMETRY. 

Ex.  81.  Give  another  demonstration  of  Case  III,  of 
Prop.  XXXVII,  by  us- 
ing   the   following   sug- 
gestions: 

SUG.    1.     Connect     C 
and  C. 

SUG.    2.     Compare 
ZACCwithZACC 

SUG.  3.   Compare  sup-  ^ 
plement  of  Z.  A  C1  C  with  Z  B  C1  C. 

SUG.  4.  Compare  supplement  of  Z.  A  C  C'  with 
Z.BCC . 

SUG.  5.  Compare  Z  B  C1  C  with  Z  B  C  C'. 

Therefore 

PROPOSITION  XXXVIII.     THEOREM. 

114.  If  two  triangles  have  two  sides  of  one  equal  to 
two  sides  of  the  other  and  the  third  sides  unequal  the 
angles  opposite  the  third  sides  are  unequal,  that  being 
greater  which  is  opposite  the  greater  third  side. 


Let  ABC  and  A'  B'  C'  represent  two  triangles  having 
A  B  equal  to  A'  B'  and  A  C  equal  to  A'  C'  and  B  C 
greater  tJian  B'  C'. 


RECTILINEAR   FIGURES.  67 

To  prove  that  the  angle  A  is  greater  than  the  angle  A' . 

SUG.  1.  If  Z  A  equals  Z  A',  how  do  B  C  and  B'  C 
compare  ? 

SUG.  2.  If  Z.  A  is  less  than  Z  A',  how  do  B  C  and 
B'  C'  compare  ? 

SUG.  3.  How,  then,  must  Z  A  compare  with  Z.  A '  ? 

Therefore 

POLYGONS. 

115.  A  polygon  is  a  portion  of  a  plane,  bounded  by 
straight  lines;  as  A  B  C  D  E. 

The  bounding  lines  are  called 
the  sides  of  the  polygon,  and 
their  sum  is  called  the  perimeter 
of  the  polygon. 

The  angles  formed  by  the  sides 
of  the  polygon  on  the  side  of  the 
enclosed  space,  are  called  the  in- 
terior angles  of  the  polygon,  as 
the  angle  B  A  E  in  the  figure. 

An  angle  formed  by  one  side  of  the  polygon  and  an 
adjacent  side  extended,  is  called  an  exterior  angle  of 
the  polygon;  as  the  angle  m  in  the  figure. 

The  vertices  of  the  interior  angles  of  a  polygon  are 
called  the  vertices  of  the  polygon. 

A  line  joining  any  two  vertices,  not  adjacent,  is  called 
a  diagonal  of  a  polygon. 

When  no  ambiguity  arises,  a  polygon  is  frequently 
read  by  naming  any  two  vertices,  not  adjacent;  as  A  D, 
B  D  etc. 


68  PLANE   GEOMETRY. 

116.  Polygons  are  classified  according  to  the  number 
of  their  sides.     The  least  number  of  sides  a  polygon  can 
have  is  three. 

A  polygon  of  three  sides  is  a  triangle. 

A  polygon  of  four  sides  is  called  a  quadrilateral. 

A  polygon  of  five  sides  is  called  a  pentagon. 

A  polygon  of  six  sides  is  called  a  hexagon.     Etc. 

117.  An   equilateral   polygon  is  a  polygon   all   of 
whose  sides  are  equal. 

An  equiangular  polygon  is  one  all  of  whose  angles 
are  equal. 

A  convex  polygon  is  a  polygon  no  side  of  which,  if 
extended,  would  enter  the  space  enclosed  by  the  perim- 
eter of  the  polygon;  as  ABCDE  in  the  figure  of 
Art.  113. 

A  concave  polygon  is  a  polygon,  two  or  more  sides 
of  which,  if  extended,  would  A 

enter  the  space  enclosed  by  the 


perimeter    of    the   polygon;    as    ^  \§. 

A  B  C  D  E  F,-  in  the  figure   at     \  / 

the  right.     If  either  A  B  or  B  C       V  / 

r  _ft  I 

be  extended  through  B  it  would 

enter  the  space  enclosed  by  the  perimeter  of  the  polygon. 
The  angle  A  B  C  in  this  figure  is  called  a  re-entrant 
angle. 

QUADRILATERAL. 

118.  Quadrilaterals  are  divided  into  classes  as  follows: 

The  trapezium,  which  has  no 
two  of  its  sides  parallel. 

The  trapezoid,  which  has  two 
sides  parallel. 

The  parallelogram,  which  has 
its  pairs  of  opposite  sides  parallel. 


RECTILINEAR   FIGURES. 


69 


C     At 


119.  The  parallel  sides  of  a  trapezoid  are  called  bases, 
the  non-parallel  sides  are  called 

legs,  and  the  perpendicular  dis- 
tance between  the  bases  is  called 
the  altitude  of  the  trapezoid. 
In  the  figure  at  the  right  A  B 
and  C  D  are  the  bases,  A  C  and 
B  D  are  the  legs,  and  A  Mis  the 
altitude  of  the  trapezoid. 

120.  Either  pair  of  opposite  sides  of  a  parallelogram 
may  be  called  bases,  but 

when  one  pair  is  selec- 
ted as  bases  the  other 
pair  must  not  be  called 
bases.  When  a  pair  of 
parallel  sides  of  a  paral- 
lelogram is  selected  as 
bases  one  of  these  sides  is  called  the  primary  base,  and 
the  other  the  secondary  base. 

The  perpendicular  distance  between  the  bases  of  a 
•parallelogram  is  called  the  altitude  of  the  parallelogram. 

Usually  the  words  lower  base  and  upper  base  are  used  instead  of 
primary  base  and  secondary  base  respectively,  but  as  geometry  does 
not  take  into  account  the  idea  of  up  and  down,  the  terms  primary  and 
secondary  are  preferable. 

121.  Parallelograms  are  classified  as  follows: 
A  parallelogram  all  of  whose 

angles  are  oblique  is  called  a 
rhomboid,  and  a  paralello- 
gram  all  of  whose  angles  are 
right  angles  is  called  a  rect- 
angle. 


B 

PARALLELOGRAM. 


70 


PLANE   GEOMETRY. 


The  distinction  between  the  different  kinds  of  parallelograms 
should  be  kept  clearly  in  mind.  Students  often  make  the  mistake  of 
drawing  the  figure  of  a  rectangle  when  proving  a  proposition  about  a 
parallelogram.  A  rectangle  is  a  parallelogram,  but  a  parallelogram 
is  not  necessarily  a  rectangle.  Every  property  of  a  parallelogram  is 
also  a  property  of  a  rectangle,  but  a  property  of  a  rectangle  may  not 
be  a  property  of  parallelograms  in  general,  and,  if  a  rectangle  is 
drawn  in  connection  with  a  proposition  about  a  parallelogram,  there 
is  danger  of  being  misled  into  applying  to  parallelograms  some  prop- 
erty which  is  true  only  of  rectangles. 


RHOMBUS.  SQUARE. 

122.  A  rhomboid  whose  sides  are  all  equal  is  called  a 
rhombus,  and  a  rectangle  whose  sides  are  all  equal  is 
called  a  square. 

PROPOSITION  XXXIX.    THEOREM. 

123.  The  opposite  sides  of  a  parallelogram  are 
equal. 


D 


Let  AB  C  D  represent  a  parallelogram. 

To  prove  that  the  opposite  sides  A  C  and  B  D  are  equal; 
also  that  the  sides  C  B  and  A  D  are  equal. 


RECTILINEAR   FIGURES.  71 

SUG.  1.  Draw  the  diagonal  A  B. 
SUG.  2.   Compare  the  As  A  B  C  and  A  B  D. 
SUG.  3.  How  then  does  A  C  compare  with  D  B,  and 
C  B  with  A  D  ? 
Therefore 

124.  COROLLARY.     The  diagonal  of  a  parallelogram 
divides  it  into  two  equal  triangles. 

PROPOSITION  XI^.    THEOREM. 

125.  If  a  quadrilateral  have  two  of  its  sides  equal 
and  parallel,  it  is  a  parallelogram. 


Let  A  C  B  D  represent  a  quadrilateral  in  which  tJie  side 
A  C  is  equal  tint!  parallel  to  the  side  B  D. 

To  prove  that  A  B  C  D  is  a  parallelogram. 

SUG.  1.  Draw  the  diagonal  A  B. 

SUG.  2.  Compare  the  As  A  C  B  and  A  D  B. 

SUG.  3.  Then  how  does  ^/  m  compare  with  ^  n  ? 
Why? 

SUG.  4.  If  Z.  m  equals  ^  n,  what  relation  does  A  D 
sustain  to  C  B  ? 

Does  the  quadrilateral  A  C  B  D  agree  with  the  defini- 
tion of  a  EU  ? 

Therefore 


72  PLANE   GEOMETRY. 

PROPOSITION  XU.     THEOREM. 

126.  A  quadrila  teral  whose  opposite  sides  are  equal 
is  a  parallelogram. 


Let  ACB  D  represent  a  quadrilateral  in  which  A  C 
equals  D  B  and  A  Z>  equals  C  B. 

To  prove  that  the  quadrilateral  A  C  B  D  is  a  parallelo- 
gram. 

SUG.  1.  Draw  the  diagonal  A  B. 
»SuG.  2.  Compare  As  A  C  B  and  A  D  B. 
SUG.  3.  How  does  Z!  m  compare  with  ^  n  ?     Why  ? 
SUG.  4.  See  Prop.  XL. 
Therefore  - 

QUERY.  How  many  and  what  statements  have  been 
made  that  may  be  used  as  definitions  of  a  parallelogram  ? 

Note. —  The  pupil  should  carefully  distinguish  between  a  property 
of  a  figure  and  a  statement  that  may  be  used  as  a  definition. 

A  definition  of  a  figure  is  such  a  description  of  it  as  serves  to  dis- 
tinguish it  from  every  other  figure.  A  property  of  a  figure  is  simply 
something  which  is  true  of  the  figure,  and  may  also  be  true  of  other 
different  figures. 


Ex.  82.  In  any  triangle  ABC,  the  angle  A  plus  twice 
the  angle  B  minus  three  times  the  angle  C  equals  110 
degrees,  and  the  angle  A  minus  twice  the  angle  B  plus 
three  time  the  angle  C  equals  90  degrees;  find  the  angles 
A,  B  and  C. 


RECTILINEAR   FIGURES.  73 

PROPOSITION  XLH.    THEOREM. 

127.  Two  parallelograms  which  have  two  sides  and 
the  included  angle  of  one  equal  to  two  sides  and  the 
included  angle  of  the  other,  each  to  each,  are  equal  in 
all  respects. 


B 


Let  AD  and  A'  J>'  represent  two  parallelograms  in 
which  A  S  equals  A'  B'9  A  C equals  A'  C'  and  the  angle 
A  equals  the  angle  A ' . 

To  prove  that  the  parallelograms  are  equal  in  all  re- 
spects. 

SUG.  1.  Place  the  £7  A  B  D  C  upon  £7  A'  B'  D'  C 
so  that  A  B  coincides  with  A'  B',  A  upon  A ',  and  B 
upon  B',  and  so  that  the  two  figures  shall  fall  upon  the 
same  side  of  A'  B'. 

SUG.  2.  What  diiection  will  A  Ctake?     Why? 

SUG.  3.  Where  will  the  point  Cfall?     Why? 

SUG.  4.  What  direction  will  C  D  take  ?     Why  ? 

SUG.  5.  Where  will  the  point  D  fall  ?     Why  ? 

Therefore 


Ex.  83.  In  proposition  XLJI,  suppose  A  C  equals 
A'  C',  CD  equals  C'  D ',  and  the  angle  C  equals  the 
angle  C'.  then  draw  the  diagonals  A  D  and  A'  D ',  and 
prove  the  proposition  in  another  way. 


74  PLANE   GEOMETRY. 

PROPOSITION  XUII.    THEOREM. 

128.  The  diagonals  of  a  parallelogram  bisect  each 
other. 


Let  K  TJ  N  M  represent  a  parallelogram,  and  L  M  and 
K  Nits  diagonals,  intersecting  at  O. 

To  prove  that  K  O  equals  O  N  and  L  O  equals  O  M. 

SUG.  Compare  A  K  O  L  with  A  M  O  N,  or  ^  L  O  N 
with  A  1C  O  M. 
Therefore 

PROPOSITION  XUV.     THEOREM. 

129.  The  opposite  angles  of  a  parallelogram  are 
equal. 


Let  A  C  B  D  be  a  parallelogram. 

To  prove  that  the  angle  CAD  equals  the  angle  C  B 
or  that  the  angle  C  equals  the  angle  D. 


RECTILINEAR   FIGURES.  75 

SUG.  1.  Draw  the  diagonal  A  B. 

SUG.  2.   Compare  ^s  m  and  n.     Give  auth. 

SUG.  3.  Compare  Zs  C  A  B  and  A  B  D.     Give  auth. 

SUG.  4.  Compare  Zs  C  A  D  and  C  B  D.     Give  auth. 

SUG.  5.  In  a  similar  manner  compare  Z!s  C  and  Z>. 

Therefore 

PROPOSITION  XLV.    THEOREM. 

130.  The  sum  of  the  interior  angles  of  any  convex 
polygon  is  equal  to  twice  as  many  right  angles  as  the 
polygon  has  sides,  minus  four  right  angles. 

B 
A 


E 
Let  A  B  C  D  E  F  represent  a  convex  polygon. 

To  prove  that  the  sum  of  the  interior  angles  of  the  poly- 
gon equals  twice  as  many  fight  angles  as  the  polygon  has 
sides,  minus  four  right  angles. 

SUG.  1 .  Connect  each  vertex  with  O,  any  point  within 
the  polygon. 

SUG.  2.  If  the  polygon  has  n  sides,  how  many  As  are 
formed  ? 

SUG.  3.  How  many  rt.  ^.s  in  the  sum  of  the  angles  of 
each  A?  How  many  in  the  sum  of  the  angles  of  all  the  As  ? 

SUG.  4.  What  is  the  sum  of  the  Zs  about  <9? 

SUG.  5.  Then,  how  many  rt.  ^s  in  the  sum  of  the  in- 
terior Z-s  of  the  polygon  ? 

Therefore 


76  PLANE   GEOMETRY. 

PROPOSITION  XI/VI.    THEOREM. 

131.  The  sum  of  the  exterior  angles  of  any  convex 
polygon  is  equal  to  four  right  angles. 


L,et  A  B  C  D  E  represent  a  convex  polygon. 

To  prove  that  the  sum  of  the  exterior  angles  a,  c,  e,  m 
and  o,  equals  four  right  angles. 

SUG.  1.  What  is  the  sum  of  the  Z.&  a  and  £?  Of 
c  and  dt  Off  and  e  ?  Etc. 

SUG.  2.  If  the  polygon  has  n  sides,  the  sum  of  all  the 
exterior  and  interior  Z^s  equals  how  many  rt.  ^.s  ? 

SUG.  3.  See  Prop.  XLV,  and  complete  the  demonstra- 
tion. 

Therefore 


PROPOSITIONS  IN  CHAPTER  I. 


PROPOSITION  I. 

At  a  given  point,  in  a  straight  line,  one  perpendicular  can  be  erec- 
ted to  that  line,  and  but  one. 

PROPOSITION  II. 

If  one  straight  line  meets  another  straight  line,  the  sum  of  the  ad- 
jacent angles  formed  equals  two  right  angles. 

PROPOSITION  III. 
If  two  straight  lines  intersect,  the  vertical  angles  formed  are  equal. 

PROPOSITION  IV. 

If  two  triangles  have  two  sides  and  the  included  angle  of  one  equal 
to  two  sides  and  the  included  angle  of  the  other,  each  to  each,  the  tri- 
angles are  equal  in  all  respects. 

PROPOSITION  V. 

If  two  triangles  have  two  angles  and  the  included  side  of  one,  equal 
to  two  angles  and  the  included  side  of  the  other,  each  to  each,  the 
triangles  are  equal  in  all  respects. 

PROPOSITION  VI. 

The  angles  opposite  the  equal  sides  of  an  isosceles  triangle  are 
equal. 

PROPOSITION  VII. 

If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight  line, 
the  distances  from  any  point  in  the  perpendicular  to  the  extremities 
of  the  line  are  equal. 

PROPOSITION  VIII. 
Any  side  of  a  triangle  is  less  than  the  sum  of  the  other  two. 


78  PLANE    GEOMETRY. 


PROPOSITION  IX. 

If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight  line, 
the  distances,  from  a  point  not  in  the  perpendicular,  to  the  extrem- 
ities of  the  line  are  unequal. 

PROPOSITION  X. 

Determine  the  locus  of  a  point  at  equal  distances  from  the  extrem- 
ities of  a  given  line. 

PROPOSITION  XI. 

Two  triangles  having  the  three  sides  of  one  equal,  respectively,  to 
the  three  sides  of  the  other,  are  equal  in  all  respects. 

PROPOSITION  XII. 

If  the  sum  of  two  adjacent  angles  equals  two  right  angles,  their  ex- 
terior sides  form  a  straight  line. 

PROPOSITION  XIII. 

From  a  point  without  a  line,  one,  and  only  one,  perpendicular  can 
be  drawn  to  the  line. 

PROPOSITION  XIV. 

Two  right  triangles  which  have  the  hypotenuse  and  a  side  of  one 
equal  to  the  hypotenuse  and  a  side  of  the  other,  are  equal  in  all  re- 
spects. 

PROPOSITION  XV. 

Two  right  triangles  which  have  the  hypotenuse  and  an  acute  angle 
of  one  equal  to  the  hypotenuse  and  an  acute  angle  of  the  other,  are 
equal  in  all  respects. 

PROPOSITION  XVI. 

Two  lines  which  are  perpendicular  to  the  same  line,  are  parallel. 

PROPOSITION  XVII. 

If  one  of  two  parallel  lines  is  perpendicular  to  a  given  line  the  other 
one  is  perpendicular  to  the  same  line. 

PROPOSITION  XVIII. 

If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate  interior 
angles  are  equal. 


RECTILINEAR    FIGURES.  79 


PROPOSITION  XIX. 

If  two  parallel  lines  are  cut  by  a  transversal,  the  corresponding 
angles  are  equal. 

PROPOSITION  XX. 

If  two  parallel  lines  are  cut  by  a  transversal,  the  interior  angles  on 
the  same  side  of  the  transversal  are  supplements  of  each  other. 

PROPOSITION  XXI. 

If  two  straight  lines  are  cut  by  a  transversal,  so  that  the  alternate 
interior  angles  are  equal,  the  lines  are  parallel. 

PROPOSITION  XXII. 

If  two  straight  lines  are  cut  by  a  transversal  so  that  the  correspond- 
ing angles  are  equal,  the  lines  are  parallel. 

PROPOSITION  XXIII. 

If  two  straight  lines  are  cut  by  a  transversal  so  that  the  interior 
angles  on  the  same  side  of  the  transversal  are  supplements  of  each 
other,  the  lines  are  parallel. 

PROPOSITION  XXIV. 

An  exterior  angle  of  a  triangle  equals  the  sum  of  the  opposite  in- 
terior angles. 

PROPOSITION   XXV. 

The  sum  of  the  interior  angles  of  a  triangle  equals  two  right  angles. 

PROPOSITION  XXVI. 

If  two  triangles  have  two  angles  of  one  equal  respectively  to  two 
angles  of  the  other,  the  third  angles  are  equal. 

PROPOSITION  XXVII. 

If  two  right  triangles  have  a  side  and  an  acute  angle  of  one  equal 
to  a  side  and  an  acute  angle  of  the  other,  the  triangles  are  equal  in  all 
respects. 

PROPOSITION  XXVIII. 

If  two  triangles  have  two  angles  and  aside  of  one  equal  respectively 
to  two  angles  and  a  corresponding  side  of  the  other,  the  triangles  are 
equal  in  all  respects. 


80  PLANE   GEOMETRY. 

PROPOSITION  XXIX. 

If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  them  are 
equal,  and  the  triangle  is  isosceles. 

PROPOSITION  XXX. 

If  two  angles  of  a  triangle  are  unequal,  the  sides  opposite  them  are 
unequal,  the  greater  side  being  opposite  the  greater  angle. 

PROPOSITION  XXXI. 

If  two  sides  of  a  triangle  are  unequal,  the  angles  opposite  them  are 
unequal,  the  greater  angle  being  opposite  the  greater  side. 

PROPOSITION  XXXII. 

If  a  perpendicular  and  oblique  lines  are  drawn  from  a  point  to  a 
given  line: 

I.  The  perpendicular  is  shorter  than  any  oblique  line. 

II.  Two  oblique  lines  which  meet  the  given  line  at  equal  distances 
from  the  foot  of  the  perpendicular,  are  equal. 

III.  Of  two  oblique  lines  meeting  the  given  line  at  unequal  dis- 
tances from  the  foot  of  the  perpendicular,  the  more  remote  is  the 
greater. 

PROPOSITION  XXXIII. 

CONVERSE  OF  PROP.  XXXII,  I.  The  shortest  line  from  a  point  to  a 
line  is  a  perpendicular  to  the  line. 

PROPOSITION  XXXIV. 

Two  equal  oblique  lines,  drawn  from  the  same  point  in  the  perpen- 
dicular to  a  given  line,  cut  off  equal  distances  from  the  foot  of  the 
perpendicular. 

PROPOSITION  XXXV. 

Two  unequal  oblique  lines,  drawn  from  the  same  point  in  the  per- 
pendicular to  a  given  line,  cut  off  unequal  distances  from  the  foot  of 
the  perpendicular,  the  longer  line  cutting  off  the  greater  distance. 

PROPOSITION  XXXVI. 

If  two  lines  be  drawn  from  a  point  in  a  triangle  to  the  extremities 
of  a  side,  the  sum  of  the  two  lines  so  drawn  is  less  than  the  sum  of  the 
other  two  sides  of  the  triangle. 


RECTILINEAR   FIGURES.  81 

PROPOSITION  XXXVII. 

If  two  triangles  have  two  sides  of  one  equal  to  two  sides  of  the  other 
and  the  included  angles  unequal,  the  remaining  sides  are  unequal, 
that  being  greater  which  belongs  to  the  triangle  having  the  greater  in- 
cluded angle. 

PROPOSITION  XXXVIII. 

If  two  triangles  have  two  sides  of  one  equal  to  two  sides  of  the  other 
and  the  third  sides  unequal,  the  angles  opposite  the  third  sides  are 
unequal,  that  being  greater  which  is  opposite  the  greater  third  side. 

PROPOSITION  XXXIX. 
The  opposite  sides  of  a  parallelogram  are  equal. 

PROPOSITION  XL. 

If  a  quadrilateral  have  two  of  its  sides  equal  and  parallel,  it  is  a 
parallelogram. 

PROPOSITION  XLI. 

A  quadrilateral  whose  opposite  sides  are  equal  is  a  parallelogram. 

PROPOSITION  XLII. 

Two  parallelograms  which  have  two  sides  and  the  included  angle  of 
one  equal  to  two  sides  and  the  included  angle  of  the  other,  each  to 
each,  are  equal  in  all  respects. 

PROPOSITION  XLIII. 
The  diagonals  of  a  parallelogram  bisect  each  other. 

PROPOSITION  XLIV. 
The  opposite  angles  of  a  parallelogram  are  equal. 

PROPOSITION  XLV. 

The  sum  of  the  interior  angles  of  any  convex  polygon  is  equal  to 
twice  as  many  right  angles  as  the  polygon  has  sides,  minus  four  right 
angles. 

PROPOSITION  XLVI. 

The  sum  of  the  exterior  angles  of  any  convex  polygon  is  equal  to 
four  right  angles. 
6— Geo. 


CHAPTER  II. 
THE  CIRCLE. 


DEFINITIONS. 

132.  A  circle  is  a  portion  of  a  plane  bounded  by  a 
curved    line    all    points    of    which    are 

equally  distant  from  a  fixed  point  within 
it. 

The  fixed  point  is  called  the  center  of 
the  circle,  and  the  bounding  line  is  called 
the  circumference  of  the  circle. 

In  higher  branches  of  mathematics  the  word 

circle  is  used  to  denote  what  is  here  defined  as  the  circumference, 
i.  e.,  the  curved  line  bounding  a  portion  of  a  plane  instead  of  that 
portion  of  a  plane  itself,  but  in  this  book  the  above  definitions  will  be 
adhered  to. 

133.  Any  straight  line  drawn  from  the  center  to  the 
circumference  is  called  a  radius.     (O  A 

is  a  radius.) 

Any  straight  line  drawn  through  the 
center,  terminated  both  ways  by  the  cir- 
cumference, is  called  a  diameter.  (C  B 
is  a  diameter.) 

From  the  definition  of  a  circle  all  radii  of  the  same 
circle  are  equal;   also,  all  diameters  of 
the  same  circle  are  equal,  each  diameter 
being  twice  the  radius. 

134.  An  arc  of  a  circle  is  any  por- 
tion of  its  circumference;  as  A  M  B. 


THE   CIRCLE.  83 


A  chord  of  a  circle  is  a  straight  line  joining  any  two 
points  on  the  circumference;  as  A  B. 
•  A  chord  of  a  circle  is  said  to  subtend  an  arc  whose 
extremities  are  the  same  as  the  extremities  of  the  chord. 

The  chord  A  B  subtends  the  arc  A  M  B,  and  also  the 
arc  A  C  D  B.  Thus,  any  chord  subtends  two  arcs 
which  together  make  up  the  whole  circumference,  but, 
when  an  arc  and  its  chord  are  spoken  of  the  smaller  of 
the  two  arcs  is  always  understood  unless  the  contrary  is 
specifically  stated. 

135.  A  secant  of  a  circle  is  any  straight  line  meeting 
the  circumference  in  two  points  and 

passing  through  at  least  one  of  these 
points;  as  C  D  E. 

A  secant  may  be  considered  as  a 
chord  extended. 

136.  A  tangent  to  a  circle  is  a  straight  line  which 

has  one  and  but  one  point  in  com-        £ ... 

mon  with  the  circle;  as  M  T  N. 

The  line  is  said  to  be  tangent  to 
the  circle,  and  the  circle  to  the  line.         I        ° 
The   common   point   is   called   the 
point  of  contact  or  point  of  tan- 
gency. 

137.  A  segment  of  a  circle  is  a  portion  of  a  circle 
bounded  by  an  arc  and  its  subtending  chord;  as  segment 
A  M  B  in  the  figure  in  article  134,  i.  e.,  the  part  of  the 
circle  included  between  the  arc  A  MB  and  the  chord  A  B. 

138.  A  sector  of  a  circle  is  a  portion  of  a  circle  bounded 
by  two  radii  and  the  intercepted  arc.     In  the  figure  in 
Art.  134,  the  sector  C  O  D  is  bounded  by  the  arc 

and  the  radii  O  C  and  O  D. 


84:  PLANE   GEOMETRY. 

The  arc  intercepted  between  two  radii  is  said  to  sub- 
tend the  angle  made  by  the  radii;  as  arc  C  D  subtends 
the  angle  COD. 

139.  A  circle  is  often  read  by  naming  the  letter  at  the 
center  of  the  circle. 

From  the  definition  of  a  circumference  it  may  be  con- 
sidered as  the  locus  of  a  point  at  a  fixed  distance  from 
the  center.  Also  the  distance  from  the  center  to  any 
point  within  the  circle  is  less  than  the  radius  and  the 
distance  from  the  center  to  any  point  without  the  circle 
is  greater  than  the  radius. 

PROPOSITION  I.    THEOREM. 

140.  Two  circles  are  equal  if  the  radius  of  one 
equals  the  radius  of  the  other. 


Let  the  figures  O  and  S  represent  two  circles  having 
equal  radii. 

To  prove  that  the  circles  O  and  S  are  equal. 

SuG.  1.  Place  the  0  O  upon  the  O  S,  with  the  center 
O  upon  the  center  5". 


THE   CIRCLE.  85 


SUG.  2.  Where  must  the  circumference  of  O  fall  with 
respect  to  the  circumference  of  ,S  ?     (See  definition  of  O.) 
Therefore 

INDIRECT  METHOD. 

SUG.  1.  Same  as  above. 

SUG.  2.  Suppose  some  part  of  circumference  of  one  O 
should  fall  outside  of  the  circumference  of  the  other  O, 
how  would  the  radii  compare? 

PROPOSITION  II.    THEOREM. 

141.  A  diameter  divides  a  circle  into  two  equal 
parts. 


Let  A  D  B  C  represent  a  circle,  and  ABa  diameter. 

To  prove  that  A  B  divides  the  circle  into  two  equal  parts. 

SUG.  1.  Revolve  the  segment  A  D  B  about  the  line 
A  O  B  into  the  plane  A  C  B. 

SUG.  2.  Where  will  the  arc  A  D  B  fall  with  respect  to 
the  arc  A  C  B  ?  (See  definition  of  O.) 

142.  A  semicircle  is  a  segment  of  a  circle  bounded 
by  a  diameter  and  the  arc  it  subtends,  as  A  D  B  in  the 
above  figure. 


86  PLANE    GEOMETRY. 

PROPOSITION  III.    THEOREM. 

143.  In  the  same  circle,  or  in  equal  circles,  equal 
angles  at  the  center  intercept  equal  arcs  at  the  cir- 
cumference. 


Let  O  and  S  represent  equal  circles,  and  let  A  O  S  and 
C  &  X>  represent  equal  angles  at  the  centers  O  and  S  respe- 
ctively. 

To  prove  that  the  arc  A  B  is  equal  to  the  arc  C  D. 

SUG.  1.  Place  O  O  upon  O  S,  so  that  the  lines  O  A 
and  O  B  fall  upon  5  C  and  5  D,  respectively.  Why  is 
this  possible  ? 

SUG.  2.  Where  will  the  arc  A  B  fall  ?     Why  ? 

Therefore 


Ex.  84.  A  parallelogram  having  one  right  angle  is  a 
rectangle. 

Ex.  85.  Prove  that  a  quadrilateral  is  a  parallelogram 
if  its  opposite  angles  are  equal. 

Ex.  86.  A  quadrilateral  is  a  parallelogram  if  its  diag- 
onals bisect  each  other. 

Ex.  87.  The  diagonals  of  a  rhombus  or  a  square  bisect 
each  other  at  right  angles. 


THE   CIRCLE.  87 


PROPOSITION  IV.    THEOREM. 

144.  Converse  of  Prop.  III.  In  the  same  circle,  or  in 
equal  circles,  equal  arcs  subtend  equal  angles  at  the 
center. 


Let  O  and  S  represent  equal  circles,  and  let  A  Band  CD 
represent  equal  arcs. 

To  prove  that  the  angle  A  O  B  equals  the  angle  C  S  D. 
SUG.  1.  Place  O  O  upon  O  5  so  that  arcs  A  B  and 
C  D  coincide.     Why  is  this  possible  ? 
Complete  the  demonstration. 
Therefore 


Ex.  88.  If  two  adjacent  sides  of  a  rectangle  are  equal 
the  figure  is  a  square. 

Ex.  89.  If  a  line  parallel  to  the  base  of  a  triangle  bi- 
sects one  side  it  bisects  the  other  side  also. 

A  B  is  bisected  at  D,  and  D  E  is  parallel 


Prove  A  E  equals  E  C. 

SUG.   Draw  D  M  \\  to  A  C. 

Ex.  90.  In  the  previous  exercise  instead 
of  drawing  D  M  parallel  to  A  C,  draw  E  M  parallel  to 
A  B,  and  prove  the  exercise. 


88  PLANE    GEOMETRY. 

PROPOSITION  V.    THEOREM. 

145.  In  the  same  circle,  or  in  equal  circles,  chords 
which  subtend  equal  arcs  are  equal. 


Let  O  and  S  represent  equal  circles,  in  which  the  arc 
A  S  equals  the  arc  €  D. 

To  prove  that  the  chord  A  B  is  equal  to  the  chord  C  D. 

SUG.  1.  Draw  the  radii  O  A,  OB,  S  C,  and  5  D. 
SUG.  2.  Compare  ^/s  O  and  ,S.     Give  auth. 
SUG.  3.  Compare  As  A  O  B  and  C  S  D.     Give  auth. 
SUG.  4.  Compare  chord  A  B  with  chord  C  D. 
Therefore 


Ex.  91.  With  the  same  construction  as  in  exercise  89, 
connect  E  and  M,  and  prove  the  exercise  without  refer- 
ence to  any  proposition  in  quadrilaterals. 

Ex.  92.  If  a  line  bisects  the  two  legs  of  a  triangle, 
prove  that  it  is  parallel  to  the  base. 

Ex.  93.  In  exercise  89,  prove  that  D  E  equals  one 
half  the  base  B  C. 


THE   CIRCLE.  89 


PROPOSITION  VI.    THEOREM. 

146.  Converse  of  Prop.  V.  In  the  same  circle,  or  in 
equal  circles,  arcs  which  are  subtended  by  equal 
chords  are  equal. 


Let  O  and  S  represent  equal  circles  in  which  the  chord 
A  jB  equals  tJie  chord  C  D. 

To  prove  that  the  arc  A  B  is  equal  to  the  ate  CD. 

SUG.  1.  Draw  radii  O  A,  OB,  S  C,  and  5  D. 

SUG.  2.  Compare  As  A  O  B  and  C  S  D.     Give  auth. 

SUG.  3.  Compare  ^s  O  and  6".     Give  auth. 

SUG.  4.  Compare  arcs  A  B  and  C  D.     Give  auth. 

Therefore 


Ex.  94.  The  diameter  of  a  parallelogram  divides  it 
into  two  equal  parts. 

A  diameter  of  a  quadrilateral  is  a  line  which  joins  the 
middle  points  of  two  opposite  sides. 

Ex.  95.  The  two  diameters  of  a  parallelogram  bisect 
each  other. 


90 


PLANE    GEOMETRY. 


PROPOSITION  VII.     THEOREM. 

147.  In  the  same  circle,  or  in  equal  circles,  two 
chords  which  subtend  unequal  arcs  are  unequal,  that 
chord  ~being  greater  which  subtends  the  greater  arc. 


Let  O  and  S  represent  equal  circles,  and  let  the  arc  A  B 
be  less  than  the  arc  C  J>. 

To  prove  that  the  chord  A  B  is  less  than  the  chord  C  D. 

SUG.  1.  Connect  the  extremities  of  the  arcs  with  their 
respective  centers. 

SUG.  2.  Place  0  O  upon  O  S,  O  upon  S,  and  A 
upon  C. 

SUG.  3.  As  arc  A  B  is  less  than  arc  C  D,  where  will 
the  point  B  fall  with  respect  to  the  arc  C  D  ? 

SUG.  4.  Compare  Z  C  S  E  with  Z  C  S  D.  Give 
auth. 

SUG.  5.  How,  then,  does  chord  C  E  compare  with 
chord  C  D,  and  hence  A  B  with  C  D  ?  Why  ? 

Therefore 


Ex.  96.  Draw  two  parallel  lines  and  a  transversal. 
Bisect  each  of  the  two  interior  angles  on  the  same  side 
of  the  transversal,  and  prove  that  the  bisectors  are  per- 
pendicular to  each  other. 


THE    CIRCLE. 


91 


PROPOSITION  VIII.     THEOREM. 

148,  Converse  of  Prop.  VII.  In  the  same  circle,  or  in 
equal  circles,  two  arcs  which  are  subtended  by  un- 
equal chords  are  unequal,  that  arc  being  greater 
which  is  subtended  by  the  greater  chord. 


Let  O  and  S  represent  equal  circles,  and  let  the  chard 
A  -B  be  less  than  tlie  cliord  C  D. 

To  prove  that  the  arc  A  B  is  less  than  the  arc  C  D. 

SUG.  1.  Draw  radii  A  O,  B  O,  C  S,  and  D  S. 

SUG.  2.  Compare  Zs  A  O  B  and  C  S  D.     Give  auth. 

SUG.  3.  Place  sector  A  O  B  upon  sector  C  S  D,  so 
that  A  O  falls  upon  C  S. 

SUG.  4.  Where  will  O  B  fall  with  respect  to  the  sec- 
tor C  S  D  ? 

SUG.  5.  Where  will  B  fall  ? 

SUG.  6.  With  A  upon  C,  and  B  located  with  respect 
to  D,  how  does  arc  A  B  compare  with  arc  C  D  ? 

Therefore 


Bx.  97.  A  B  C  is  an  isosceles  triangle  whose  base  is 
B  C  and  whose  vertex  is  A:  Extend  B  A  through  A  to 
Q,  making  A  O  equal  to  B  A.  Connect  O  and  C.  Prove 
that  O  C  is  perpendicular  to  B  C. 


92 


PLANE   GEOMETRY. 


PROPOSITION  IX.    THEOREM. 

A  radius  which  is  perpendicular  to  a  chord 
bisects  the  chord  and  its  subtended  arc. 


Let  O  Dbea  radius  perpendicular  to  the  chord  L  M. 

To  prove  that  L  E  equals  E  M,  and  that  arc  L  D  equals 
arc  D  M. 

SuG.  Compare  the  As  L  £  O  and  M  E  O. 
Complete  the  demonstration. 
Therefore 


Ex.  98.  If  A  is  the  vertex  and  B  C  the  base  of  an  isos- 
celes triangle  ABC,  and  if  from  any  point  D  in  the  side 
A  B  a  line  be  drawn  perpendicular  to  the  base  and  meet- 
ing C  A  extended  at  E,  prove  that  the  triangle  A  D  E 
is  isosceles. 

Ex.  99.  If  a  perpendicular  be  drawn  from  the  vertex 
of  the  right  angle  of  a  right-angled  tri-  ^ 

angle  to  the  hypotenuse,  prove  that  the 
two  triangles  formed  are  mutually  equi- 
angular to  the  original  triangle. 

Prove  that  As  A  B  M  and  A  C  M  are 
mutually  equiangular  to  A  A  B  C. 


THE    CIRCLE. 


PROPOSITION  X.     THEOREM. 

150.  A  line  perpendicular  to  a  chord  at  its  middle 
point  passes  through  the  center. 


Let  A  B  represent  a  cliord,  and  C  D  a  perpendicular  t& 
A  B  at  its  middle  point  C. 

To  prove  that  C  D  passes  through  the  center  of  the  circle. 

SUG.  1.  From  0,  the  center,  of  the  circle,  drop  a  J_  to 
the  chord  A  />. 

SUG.  2.  Where  will  C  O  lie  with  respect  to  C  D  ? 
Why? 

Therefore 

151.  COROLLARY  I.     If  two  circles  intersect,  the  line 
joining  their  centers  is  perpendicular  to  their  common 
chord  at  its  middle  point. 

SUG.  Erect  a  _L  to  the  common  chord  at  its  middle 
point  and  extend  both  ways. 

152.  COROLLARY  II.     If  two  equal  circles  intersect, 
the  common  chord  bisects  at  right  angles  the  line  join- 
ing their  centers. 


PLANE    GEOMETRY. 


PROPOSITION  XI.     THKORKM. 

153.  In  the  same  circle,  or  in  equal  circles,  equal 
chords  are  equally  distant  from  the  center,  and  of  two 
unequal  chords  the  greater  is  nearer  the  center. 


I.  Let  A  B  and  C  D  represent  equal  chords  in  the 
equal  circles  O  and  S,  and  O  M  and  S  N  their  respective 
distances  from  the  centers  O  and  S. 

To  prove  that  O  M  equals  S  N. 

SUG.  1.   Draw  the  radii  O  B  and  5  D. 
SUG.  2.  Compare   A  O  MB    with    A  S  N  D.      Give 
auth. 
SUG.  3.  What  is  your  conclusion  about  the  equality  of 


II.  In  the  circle  O  let  A  B  and  A  E  represent  unequal 
chords,  A  B  being  greater  tlian  A  E,  and  O  M  and  O  P 
their  respective  distances  from  the  center  O. 

To  prove  that  O  M  is  less  than  O  P. 

SUG.  1.  Compare  arc  A  E  with  arc  A  B.     Prop.  VIII. 
SUG.  2.  Where  does  E  lie  with  respect  toA£?    Why  ? 
SUG.  3.  L,et  F  represent  the  intersection  of  P  O  and 
A  B.     Why  will  they  intersect  ? 


THE   CIRCLE. 


95 


SUG.  4.   Compare  O  M  with  O  F,  O  F  with  O  P,  and, 
finally,  O  M  with  O  P. 

SUG.  5.  Compare  5*  N  with  O  P. 
Therefore 

PROPOSITION  XII.     THEOREM. 

1 54.  A  straight  line  perpendicular  to  the  radius  of 
a  circle  at  its  extremity  is  tangent  to  the  circle. 


Let  O  .B  represent  a  straight  line  perpendicular  to  the 
radius  O  S  at  its  extremity  O. 

To  prove  that  O  B  is  tangent  to  the  circle. 

SUG.  1.  The  point  O  is  common  to  the  circle  5  and  the 
line  OB. 

SUG.  2.  I^et  M  represent  any  other  point  in  O  B. 
Draw  S  M. 

SUG.  3.  Compare  5*  O  and  S  Mm  respect  to  length. 

SUG.  4.  Where,  then,  is  M  with  respect  to  the  circle  ? 
See  Art.  139. 

Therefore 


96  PLANE   GEOMETRY. 

PROPOSITION  XIII.     THEOREM. 

155.  Converse  of  Prop .  XII  If  a  straight  line  is  tan- 
gent  to  a  circle  the  radius  meeting  it  at  the  point  of 
contact  is  perpendicular  to  it. 


Let  O  B  represent  a  tangent  to  the  circle  89  O  the  point 
of  contact,  and  S  O  the  radius  drawn  to  the  point  of  con- 
tact. 

To  prove  that  S  O  is  perpendicular  to  O  B. 

SUG.  1.  Draw  6*  M,  any  other  line  from  6*  to.O  B. 
SUG.  2.  Compare  6*  O  and  S  M  in  respect  to  length. 
SUG.  3.  See  article  108. 
Therefore 

156.  CORONARY.  At  any  point  in  a  circumference, 
one,  and  only  one,  tangent  can  be  drawn. 


Ex.  100.  In  the  figure  at  the 
right,  A  E  and  B  C  are  paral- 
lel, and  M  O  is  a  transversal. 
B  O  bisects  the  angle  A  O  M, 


and  CO  bisects  the  angled  O  M.     B  '  JM          C 

Prove  that  B  M  equals  M  C. 


THE    CIRCLE.  97 


PROPOSITION  XIV.     THEOREM. 

157.  Arcs  of  a  circle  intercepted  by  parallel  chords 
are  equal. 


Let  A  B  and  C  Z>  represent  two  parallel  lines  intercept- 
ing the  arcs  A  C  and  B  D. 

To  prove  that  arc  A  C  equals  arc  B  D. 

SUG.  1.  Drop  a  J_  from  O  to  CD,  and  extend  it  to 
meet  the  circumference  at  M. 

SUG.  2.  How  is  O  M  related  to  A  B  ?     See  Art.  85. 

SUG.  3.  See  Prop.  IX,  and  complete  the  demonstra- 
tion. 

Therefore 


Ex.  101.  If  one  of  the  equal  sides  of  an  isosceles  tri- 
angle be  extended  through  the  vertex  and  the  exterior 
angle  formed  be  bisected,  prove  that  the  bisector  is  par- 
allel to  the  base. 

Ex.  102.  If  the  diagonals  of  a  parallelogram  are  equal, 
prove  that  the  parallelogram  is  a  rectangle. 

Ex.  103.  The  bisectors  of  the  interior  angles  of  a  par- 
allelogram form  a  rectangle. 
7-Geo. 


98  PLANE    GEOMETRY. 

PROPOSITION  XV.     THEOREM. 

158.   Through  three  points,  not  in  the  same  straight 
line,  one  circumference,  and  only  one,  is  possible. 


A*  •B 


Let  A,  B  and  C  represent  three  given  points  not  in  the 
same  straight  line. 

To  prove  that  through  A,  B  and  C  one  circumference, 
and  but  one^  is  possible. 

SUG.  1.  What  is  the  locus  of  a  point  equally  distant 
from  A  and  C?  Give  auth. 

SUG.  2.  What  is  the  locus  of  A  point  equally  distant 
from  A  and  B  ?  Give  auth. 

SUG.  3.  Will  these  two  loci  intersect  ?     Why  ? 

SUG.  4.  Is  there  a  point  equally  distant  from  .A,  B, 
and  C? 

SUG.  5.  Can  a  circumference  pass  through  the  points 
A,  #and  C? 

SUG.  6.  Can  there  be  more  than  one  such  circumfer- 
ence ?  Why  ? 

Therefore 


THE   CIRCLE.  99 


MEASUREMENT. 

159.  Anything  which  can  be  measured  by  a  unit  of 
the  same  kind  is  called  quantity. 

The  quantities  used  in  geometry  are  the  geometric 
magnitudes,  viz. ,  lines,  surfaces  and  solids. 

To  measure  a  quantity  is  to  find  out  how  many  times 
it  contains  another  selected  quantity  of  the  same  kind, 
called  the  unit  of  measure. 

In  every  day  experience,  the  unit  of  measure  is  a  standard  ac- 
cepted by  general  consent;  as  a  foot,  a  square  yard,  a  ton,  a  cord,  etc. 

160.  The  measure  of  a  quantity  is  the  number  which 
expresses  how  many  times  the  unit  of  measure  is  con- 
tained in  the  given  quantity. 

A  quantity  is  completely  expressed  when  the  measure 
and  unit  of  measure  are  both  expressed;  as  three  yards. 
In  this  example  three  is  the  measure,  a  yard  is  the  unit 
of  measure,  and  three  yards  is  the  quantity  completely 
expressed. 

161.  The  ratio  of  two  quantities  is  the  number  of 
times  the  first  contains  the  second;  i.  e.,  the  measure  of 
the  first  regarding  the  second  as  the  unit  of  measure;  or, 
having  measured  both  quantities  by  the  same  unit,  the 
quotient  of  the  measure  of  the  first  divided  by  the  meas- 
ure of  the  second. 

The  ratio  of  line  A   to  line  B  is  the  A 

number  of  times  A  contains  B,  which 
may^be   determined   by  laying   off  B  B 

upon  A.      Or,   if  A    and  B  b^  meas- 
ured  by   same   unit   mt    the    ratio   of  m 
A  to  B  is   the  number  of  times   the 
measure  of  A  contains  the  measure  of  B.     Suppose  m 


100  PLANE    GEOMETRY. 

is  contained  c  times  in  A,  and  d  times  in  B,  then  the 
ratio  of  A  to  B  equals  c  divided  by  d>  which  may  be  ex- 

c 
pressed  as  -7- 

As  ratio  plays  an  important  part  in  subsequent  geometry,  it  is  im- 
portant that  the  pupil  have  clear  and  definite  ideas  of  what  a  ratio 
actually  is,  and  the  relation  it  sustains  to  the  subject  of  division. 

From  the  definition  of  the  ratio  of  two  quantities  as  the  number  of 
times  the  first  contains  the  second,  it  follows  immediately  that  a  ratio 
can  exist  only  between  quantities  of  the  same  kind,  and  also  that  the 
ratio  of  two  quantities  is  always  a  pure  number'.  For  example,  6  ft. 
contains  3  ft.  twice,  hence  the  ratio  of  6  ft.  to  3  ft.  is  2, —  not  2  ft.» 
but  simply  2. 

Again,  division  has  been  defined  as  the  process  of  finding  how 
many  times  one  quantity  is  contained  in  another.  Now,  it  is  clear 
that,  with  this  conception  of  division,  the  quotient  is  identical  with 
the  ratio  of  two  quantities,  as  above  defined.  But  there  are  two  kinds 
of  division;  the  one  just  described  maybe  called  the  division  of  meas- 
urement; the  other  is  the  division  of  separation,  and  means  the  proc- 
ess of  dividing  or  separating  a  quantity  into  a  certain  number  of  equal 
parts.  For  example,  6  ft.  may  be  divided  into  three  equal  parts,  and 
each  part  will  be  2  ft.  In  the  division  of  measurement  the  divisor  is 
a  quantity  of  the  same  kind  as  the  dividend,  and  the  quotient  is  a 
pure  number;  but,  in  the  division  of  separation,  the  divisor  is  a  pure 
number,  and  the  quotient  is  a  quantity  of  the  same  kind  as  the  divi- 
dend. Now,  the  ratio  of  two  quantities  identifies  itself  with  the  quo- 
tient only  in  the  division  of  measurement. 

As  the  ratio  of  two  quantities  is  the  quotient,  in  one  kind  of  divis- 

A 

ion,  the  ratio  of  A  to  B  is  often  written  in  the  fractional  form,  — .     It 

B 

is  very  important  to  keep  constantly  in  mind  the  fact  that  A  and  B 
are  quantities  of  the  same  kind,  and  that  the  ratio  -^  is  zpure  number. 
The  case  considered  above,  in  which  A  and  B  are  lines,  and  each 
is  measured  by  the  the  same  third  line  m,  gives  ^=^-  Jt  wil1  be  seen 
here  that  A  and  B  are  geometric  quantities,  and  c  and  </are  numbers. 

162.  Two  quantities  are  said  to  be  commensurable 
when  each  can  be  exactly  measured  by  the  same  unit 


THE   CIRCLE.  101 


i.  e.,  when  their  ratio  can  be  expressed  as  a  fraction 
whose  numerator  and  denominator  are  whole  numbers. 

PROPOSITION  XVI.     PROBLEM. 

163,   To  find  the  greatest  common  unit  of  measure 
of  two  given  lines. 


A 


•D 


Let  A  B  and  C  D  represent  two  commensurable  lines, 
A  B  being  greater  tJian  C  D. 

To  find  the  greatest  common  unit  of  measure  of  A  B 
and  CD. 

SUG.  Employ  the  method  for  finding  the  greatest  com- 
mon divisor  of  numbers,  by  division,  viz. : 

The  greatest  common  unit  cannot  be  greater  than  C  D, 
the  smaller  quantity.  If  C  D  is  exactly  contained  in  A  B, 
C  D  is  the  greatest  common  unit.  I,ay  off  C  D  upon  A  B 
as  many  times  as  possible.  There  may  be  a  remainder 
E  B.  The  greatest  common  unit  of  C  D  and  A  B  must 
be  exactly  contained  in  this  remainder  E  B>  and  hence 
can  be  no  greater  than  E  B.  If  E  B  is  exactly  contained 
in  C  D,  then  E  B  is  the  greatest  common  unit;  for  if  it  is 
contained  in  C  D  it  is  contained  in  any  number  of  times 
C  D,  and  hence  in  A  E  and  also  in  A  E  plus  E  B.  I^ay 
off  E  B  upon  C  D  as  many  times  as  possible.  If  there  is 
a  remainder  proceed  as  before.  When  a  remainder  is 
obtained  which  is  exactly  contained  in  the  last  preceding 
divisor,  it  is  the  greatest  common  unit. 

The  student  should  give  the  reason  for  each  statement 
in  the  above  process. 


102 


PLANE   GEOMETRY. 


PROPOSITION  XVII.     PROBLEM. 

1 64.  Given  the  greatest  common  unit  of  measure  of 
two  quantities  to  find  their  ratio. 

SUG.  See  definition  of  ratio. 

165.  Two  quantities  are  said  to  be  incommensurable 
when  it  is  impossible  to  exactly  measure  both  of  them 
by  the  same  unit,  i.  *?.,  to  express  their  ratio  as  a  frac- 
tion whose  numerator  and  denominator  are  whole  num- 
bers.    The  side  and  diagonal  of  a  square  may  be  shown 
to  be  incommensurable,  as  follows: 


B 


Let  A  C  represent  a  square,  and  B  D  the  diagonal. 

To  find  whether  there  is  a  common  unit  of  measure  of  the 
side  and  diagonal. 

Upon  the  diagonal  B  D  lay  off  D  E  equal  to  the  side 
DC,  as  in  Prop.  XVI. 

There  is  a  remainder  B  E,  and  the  greatest  common 
unit  of  measure  of  D  C  and  D  B  is  the  greatest  common 
unit  of  measure  of  B  E  and  DC,  or  of  B  E  and  B  C. 

At  E,  draw  E  F,  _L  to  the  diagonal  B  D. 

In  the  rt.  &BEF\hzZ_FBE=  half  a  rt.  Z,  and 
hence  the  Z  BFE  =  half  art.  Z.  Hence  B  E  =  E  F. 


THE   CIRCLE.  103 


Again,  if  a  line  D  F  were  drawn,  the  two  rt.  As  D  C  F 
and  DBF  would  have  the  same  hypotenuse,  and  the 
sides  D  C  and  D  E  equal.  Hence  E  F  =  F  C. 

Our  problem  now  is  to  find  the  greatest  common  unit 
of  measure,  if  one  exists,  of  B  E  and  B  C. 

Upon  the  diagonal  F  B  lay  off  F  M  =  B  E.  Then  as 
C  Fand  F  M  are  each  equal  to  B  E,  it  is  clear  that  B  E 
is  contained  twice  in  B  C  with  the  remainder  B  M.  The 
greatest  common  unit  of  measure,  if  one  exists,  of  B  E 
and  B  C  is  the  greatest  common  unit  of  measure  of  B  M 
and  B  E. 

At  M  draw  M  R,  _L  to  the  diagonal  B  F.  In  the  same 
manner  that  it  was  proved  that  B  E  =*  E  F '  =  F  C  it 
may  also  be  shown  that  B  M  —  M  R  =  RE. 

Then  B  M  is  the  side  of  a  square  of  which  B  R  is  the 
diagonal. 

If  we  now  lay  off  on  B  E  a  line  equal  to  B  M,  the  con- 
ditions are  the  same  as  when  B  E  was  laid  off  on  B  C. 
However  far  the  process  is  continued  it  is  evident  that 
each  remainder  becomes  the  side  of  a  new  square,  and 
the  last  preceding  divisor  is  divided  into  two  parts,  one 
of  which  is  the  diagonal  of  the  square,  and  the  other  a 
side.  Hence,  at  no  time  in  the  process  will  the  division 
be  exact,  and  therefore  no  common  unit  of  measure  exists 
for  the  diagonal  and  side  of  a  square.  Hence  they  are 
incommensurable. 

166.  The  ratio  of  two  incommensurable  quantities  is 
called  an  incommensurable  ratio,  or  an  incommensurable 
number.  The  ratio  of  the  diagonal  and  side  of  a  square 

has  been  computed  to  equal          - 


104  PLANE   GEOMETRY. 

Note.  —  An  incommensurable  number  must  not  be  looked  upon  as 
an  inexact  number.  If  the  side  of  a  square  is  one  foot  the  diagonal 
is  ]/  2  feet,  but  this  diagonal  is  a  perfectly  definite  length. 

167-  A  constant  is  a  quantity  whose  value  is  fixed. 

A  variable  is  a  quantity  which,  under  the  conditions 
imposed  upon  it,  may  assume  an  indefinite  number  of 
values.  For  example,  the  distance  from  a  railway  sta- 
tion to  a  moving  train  of  cars  is  a  quantity  which  has 
one  value  at  one  time,  another  value  a  minute  later,  and 
still  another  value  another  minute  later,  and  so  on. 

168,  When  a  variable  so  changes  that  it  continually 
approaches  sonfe  fixed  quantity  which  it  cannot  reach, 
but  from  which  it  may  be  made  to  differ  by  an  amount 
less  than  any  assigned  quantity,  however  small,  the  fixed 
quantity  is  called  the  limit  of  the  variable,  and  the  va- 
riable is  said  to  approach  its  limit. 

Note. —  By  the  above  definitions  it  will  be  noticed  that  not  all  va- 
riables have  limits;  but  those  considered  in  the  following  pages  do 
have  limits. 

169.  ILLUSTRATIONS.     One  mih  of  a  given  line  is  a  va- 
riable which  approaches  zero  as  a  limit  if  m  is  made  to 
increase  indefinitely,  and  a  minus  one  mth  of  a  is  a  vari- 
able which  approaches  a  as  a  limit  if  m  is  made  to  in- 
crease indefinitely. 

If  a  point  be  made  to  move  along  a  given  line,  the  dis- 
tance from  a  fixed  point  to  the  moving  point  is  a  vari- 
able. This  variable  may  or  may  not  have  a  limit.  If 
such  conditions  be  imposed  upon  the  moving  point  that 
during  a  second  it  moves  along  half  the  length  of  the 
line,  the  next  second  half  the  remaining  distance,  the 
next,  half  the  remaining  distance,  etc.,  the  length  of  the 
given  line  is  the  limit  of  the  variable. 


THE    CIRCLE.  105 


The  pupil  should  remember  that  the  limit  and  the  va- 
riable are  always  the  same  kind  of  quantities;  as  in  the 
example  just  given,  the  limit  and  variable  are  both  dis- 
tances. 

PROPOSITION  XVIII.    THEOREM. 

170.  //  two  variables  are  always  equal  as  they  ap- 
proach tlieir  limits,  their  limits  are  equal. 


O  O 


D 

Let  A  m  and  C  n  represent  two  variables,  and  A  JB  and 
C  D  tlieir  respective  limits. 

To  prove  that  A  B  equals  C  D. 

Lay  off  A  B  upon  C  D.     Let  A  B  equal  C  O. 

By  hypothesis  A  m  and  C  n  are  always  equal,  i.  e. : 
A  m  =  C  n,  A  m'  =  Cn',  etc. 

But  as  A  m  approaches  A  B,  or  its  equal  C  O,  as  its 
limit,  and  C  n  approaches  C  D  as  its  limit,  there  must 
come  a  time,  if  C  O  is  greater  or  less  than  C  D,  when 
the  value  A  m  is  not  equal  to  C  n,  which  is  contrary  to 
the  hypothesis.  Hence  C  O,  or  its  equal  A  B,  must 
equal  CD. 

Therefore,  if  two  variables  are  always  equal  as  they 
approach  their  limits,  their  limits  are  equal. 


Ex.  104.  If  the  middle  points  of  the  sides  of  a  quad- 
.  rilateral  be  joined  in  order,  the  figure  formed  is  a  paral- 
lelogram. 

SuG.  Draw  the  diagonals  of  the  given  quadrilateral. 


106 


PLANE    GEOMETRY. 


PROPOSITION  XIX.     THEOREM. 

171.  In  the  same  circle,  or  in  equal  circles,  two 
angles  at  the  center  have  the  same  ratio  as  the  arcs 
which  they  intercept  at  the  circumference. 


D 


Let  O  and  S  represent  two  equal  circles,  A  O  U  and 
C  S  -D  two  angles  at  their  centers9  and  A  JB  and  C  2>  the 
arcs  which  they  intercept  at  the  circumference. 

.,    .  Z.  A  O  B  cere  A  B 

To  prove  that  ^r-^  equals  -^^. 

There  are  two  cases. 

CASE  I.  When  the  angles  A  O  B  and  C  S  D  are  com- 
mensurable. 

SuG.  1.  I/et  £  m  represent  the  common  unit  of  meas- 
ure for  the  two  angles. 

SuG.  2.  If  Z.  m  is  contained  5  times  in  Z!  A  O  B  and 

4  times  in  Z!  C  S  D,  what  is  the  ratio      .  r  ?  n  e(lual 

to? 

SuG.  3.  Extend  the  lines  of  division  of  the  ^/s  to 
the  arcs.  How  do  arcs  A  E,  E  F,  C  I,  I K,  etc. ,  com- 
pare ?  Why  ? 

SUG.  4.  How  does  the  number  of  arcs  in  arc  A  B  com- 
pare with  the  number  of  Z.$>  in  ^  A  O  B  ? 


THE    CIRCLE.  107 


How  does  the  number  of  arcs  in  arc  C  D  compare  with 
the  number  of  ^.s  in  Z.  C  S  Z?? 

Sue.  5.  Then  the  ratio   -^-=  equals  what? 

arc  CD 

SUG.  6.  Compare  the  ratio  of  the  ^s  with  the  ratio  of 
the  arcs. 

Therefore 

CASE  II.      When  the  angles  A  O  B  and  C  S  D  are  in- 
commensurable. 


D 


Let  the  angles  A  O  B  and  C  8  D  be  incommensurable. 

SUG.  1.  Take  any  unit  of  measure  of  one  of  the  ^.s,  as 
of  Z.  C  S  D,  and  apply  this  unit  to  the  Z  A  O  B.  There 
mnst  be  a  remainder.  Why  ? 

Let  this  remainder  be  represented  by  /.MOB. 

SUG.  2.  The  Zs  A  O  M  and  COD  are  commensur- 
able. Why  ? 

SUG.  3.  Compare  the  ratio     >  ^r-  with  the   ratio 

£.  C  o  U 

arc  AM     ^. 

„  ,~.     Give  auth. 
arc   CD 

SUG.  4.  By  taking  the  unit  of  measure  of  the  Z.  C  SD 
smaller  and  smaller  continually,  the  remainder,  viz.,  the 
Z.  M  O  B,  may  be  made  as  small  as  we  please,  i.  e.,  it 
may  be  made  to  approach  zero  at  its  limit.  Can  it  be 
made  to  disappear  entirely  ?  Why?  (See  Sug.  1.) 


108  PLANE    GEOMETRY. 

SUG.  5.  The  Z  A  O  M  is  a  variable.      Why  ?     (See 
definition  of  variable.) 

SUG.  6.  Is  Z.  C  S  D  a  variable  or  a  constant  ? 

^  A.  O  M 

SUG.  7.  Is  the  ratio  -,   _  ?  _    a  constant  or  a  vari- 
able ?     Why  ? 

SUG.  8.  What  is  the  limit  of   ^  ^  c  ff  *•  *••  toward 

^L.  C   o  Xx 

,          ,.   .    Z.A  O  M 
what  ratio  is     .  —  approaching  ? 

SUG.  9.  Is  the  ratio  ^-=j  a  constant  or  a  variable  ? 

arc  CD 

Why? 

fly/*  ^\   H/F 

SUG.  10.  What  is  the  limit  of  -   ^^    z.  <?.,  toward 

«r<:  CZ^> 

.    arc  A  M 

what  ratio  is 7=r^r  approaching  r 

arc  CD 

^.  A  O  M 
SUG.  11.  Compare  the  two  variables  en    an<^ 

^_  C  o  D 
/y  y/"  /4   1W 

„  _      as  they  approach  their  limits.     (See  Sug.  3.) 
arc  CD 

SUG.  12.  Compare  their  limits.     (Prop.  XVIII.) 
Therefore 


Ex.  105.  The   diameters   of  any  quadrilateral   bisect 
each  other. 

SUG.  See  definition  of  diameter  in  exercise  94. 

Ex.  106.  The  sum  of  the  angles  at  the  vertices  of  a 
five-pointed   star    is    equal   to   two   right 
angles.     See  figure  at  the  right. 

Ex.  107.  If  a  line  be  drawn  in  a  trape- 
zoid,  bisecting  one  of  the  legs  and  parallel 
to  the  bases,  prove  that  it  bisects  the  other 
leg  also. 


THE   CIRCLE. 


109 


MODEL. 


PROPOSITION  XIX.    THEOREM. 

172.  In  the  same  circle,  or  in  equal  circles,  two 
angles  at  the  center  have  the  same  ratio  as  the  arcs 
which  they  intercept  at  the  circumference. 


Let  O  and  S  represent  two  equal  circles,  A  O  B  and 
C  S  D  two  angles  at  tJieir  centers,  and  A  B  ana  C  I>  the 
arcs  which  they  int<  rccpt  at  the  circumference. 

.,    .  Z-  A  O  B  ,    arc  A  B 

To  brove  that     ,  ^  0   ^  equals 


CS  D 

There  are  two  cases. 


arc  C 


CASE  I.      When  the  angles  A  O  B  and  C  S  D  are  com- 
mensurable, 

Let  /.  m  represent  the  common  measure  of  Z.S  A  O  B 
and  C  S  D. 

If  m  is  contained  5  times  in  Z.  A  O  B  and  4  times  in 


=      -     (See  definition  of 


Z.  CS  D,  the  ratio 

/      (-     O     - 

ratio.) 

Extend  the  sides  of  the  ^s  to  the  arcs;  then  arcs  A  E> 
E  F,C  I,  I  K,  etc.,  are  all  equal. 


110 


PLANE   GEOMETRY. 


Hence,  the  ratio 


(In  the  same  circle,  or  in  equal  circles,  equal  angles  at  the  center  in- 
tercept equal  arcs.) 

There  are  5  equal  arcs  in  arc  A  B,  and  4  in  arc  C  Dt 
for  each  angle  at  the  center  intercepts  an  arc. 
arc  A  B  _  5 
arc  CD  ==  4' 

Z  A  O  B        arc  A  B 

Therefore,        .   _  =  -  -. 

Z.  C  S  D        arc  C  D 

Therefore,  when  the  angles  at  the  centers  are  commen- 
surable they  have  the  same  ratio  as  their  intercepted 
arcs. 

CASE  II.  When  the  angles  A  O  B  and  C  S  D  are  in- 
commensurable. 


B 


0 


Let  the  angles  A  O  S  and  C  S  D  be  incommensurable. 

Take  a  unit  of  measure  of  one  of  the  Zs,  as  of  /.  C  S  D, 
and  apply  this  unit  to  the  Z.  A  O  B.  Since  the  Zs  are 
incommensurable,  this  unit  will  be  contained  in  Z_  A  OB 
a  certain  number  of  times  with  a  remainder,  as  Z  MOB. 

Since  the  Zs  A  O  M  and  C  S  D  have  a  common  unit 
of  measure,  they  are  commensurable,  also  the  arcs  A  M 
and    CD  are   commensurable,    and   hence,  by  case   I, 
Z  A  O  M  _  arc  A  M 
Z  CS  D    "S  arcCD  ' 


THE  CIRCLE.  Ill 


Now,  by  taking  the  unit  of  measure  of  the  ,/  C  S  D 
smaller  and  smaller  continually,  the  remainder,  viz.,  the 
^L  M  O  B,  which  is  always  less  than  the  unit,  may  be 
made  as  small  as  we  please,  but  cannot  be  made  to  dis- 
appear entirely,  for  then  the  Zs  A  O  B  and  C  S  D  would 
be  commensurable.  Hence,  the  remainder,  MOB,  ap- 
proaches zero  as  a  limit,  and,  therefore,  the  Z  A  O  M 
approaches  the  /LA  O  B  as  a  limit. 

But  the  Z.  C  S  D  is  a  constant,  and  hence  the  ratio 

Z  A  O  M 

/  r  c  ri~    1S   a   variable   which   approaches    the   ratio 
2—  C  o  JJ 

AAOB 


arc  A  M 

Again,  the  ratio   --  ^  _      is   a  variable   which  ap- 
arc  CD 

.     arc  A  B  ,.    . 

proaches  the  ratio  -  —  F-^T  as  a  limit. 
arc  CD 

^AOM        .    arc  A  M 

Hence,  as      .„„„.-    and   -       _  _    are  two  variables 
Z  C  S  D  arc  C  D 

which  are  always  equal  as  they  approach  their  limits, 
their  limits  must  be  equal  (by  Prop.  XVIII).     That  is, 
Z  A  OB        arc  A  B 
Z  CS  D~  !=  arc  C  D' 

Therefore,  in  the  same  circle,  or  in  equal  circles,  two 
angles  at  the  center  have  the  same  ratio  as  the  arcs 
which  they  intercept  at  the  circumference. 

173,  SCHOLIUM  I.  A  degree,  which  has  already  been 
denned  as  ^  of  a  right  angle,  or,  what  is  the  same  thing, 
•5^-ff  of  the  whole  angular  magnitude  about  a  point,  is 
often  taken  as  a  standard  unit  angle;  and  the  arc  it  in- 
tercepts on  the  circumference,  also  called  a  degree,  or 
-gfa  of  the  circumference,  is  often  taken  as  a  standard 
unit  arc. 


112  PLANE    GEOMETRY. 

174.  SCHOLIUM  II.     By   the   proposition   it   will   be 
seen  that  the  number  of  times  the  unit  angle  is  contained 
in  the  given  angle  at  the  center  is  the  same  as  the  num- 
ber of  times  the  unit  arc  is  contained  in  the  arc  inter- 
cepted by  the  given  angle.     Hence,  we  say,  briefly,  that 
an  angle  at  the  center  is  measured  by  its  intercepted  arc. 

175.  SCHOLIUM  III.     A  right  angle  which  contains 
90  degrees  (written  90°),  is  also  often  used  as   a  unit 
angle  in  estimating  the  size  of  other  angles. 

176.  An  inscribed  angle  is  one  whose  vertex  is  in 
the  circumference,   and  whose  sides  are  chords   of  the 
circle.     The  angle  D  E  F  is  an  inscribed 

angle.  An  angle  is  inscribed  in  a  segment 
of  a  circle  when  its  vertex  is  in  the  arc  of 
the  segment  and  its  sides  meet  the  ex- 
tremities of  the  chord  of  the  arc.  The 
angle  D  £  F  is  inscribed  in  the  segment 
DBF. 

PROPOSITION  XX.    THEOREM. 

177.  An  inscribed  angle  is  measured  by  one  half 
its  intercepted  arc. 


Let  A  be  an  inscribed  angle,  whose  sides  intercept  the 
arc  B  C. 


THE    CIRCLE. 


113 


To  prove  that  the  angle  A  is  measured  by  one  half  the 
arc  B  C. 

There  are  three  cases. 

CASE  I.      When  one  side  of  the  angle  passes  through  the 
center  of  the  circle. 

SUG.  1.   Connect  O,  the  center  of  the  circle,  with  C. 
SUG.  2.   Compare  Z  A  with  Z.  O  C  A.     Give  auth. 
SUG.  3.   Compare  /.BO  Cwith  Z  A.     Give  auth. 
SUG.  4.   By  what  arc  is  Z  B  O  C  measured  ?     Why  ? 
SUG.  5.  Then,  by  what  part  of  the  arc  B  C  is  the  Z  A 
measured  ? 


CASK  II. 
angle. 


When  the  center  lies  between  the  sides  of  the 


SUG.  1.  Through  the  vertex  A  draw  a  diameter  A  M. 

SUG.  2.  By  what  arc  is  the  /.BAM  measured? 
Why? 

SUG.  3.  By  what  arc  is  the  /.CAM  measured? 
Why? 

SUG.  4.  Then,  by  what  arc  is  the  Z.  B  A  C  measured  ? 

8— Geo. 


114  PLANE    GEOMETRY. 

CASE  III.      When  both  sides  of  the  angle  are  on  the  same 
side  of  the  center. 


SUG.  1.  Through  the  vertex  A  draw  a  diameter  A  M. 
SUG.  2.  By  what  arc  is  the  /.MAC  measured? 
SuG.  3.  By  what  arc  is  the  /_  M  A  B  measured  ? 
SUG.  4.  Then,  by  what  arc  is  the  ^  B  A  C  measured? 
SuG.  5.   Compare   the   last   suggestion   with   Sug.  4, 
case  II;  and  Sug.  5,  case  I. 
Therefore 

PROPOSITION  XXI.    THEOREM. 

178.  An  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  one  half  the  intercepted  arc. 

A B 


Af 
Let  A  B  represent  a  tangent,  and  AC  a  chord. 

To  prove  that  the  angle  B  A  C  is  measured  by  one  half 
the  arc  ADC. 


THE   CIRCLE.  115 


SUG.  1.  Draw,  through  A,  the  diameter  A  M. 

SUG.  2.  By  what  arc  is  the  Z  MA  C  measured? 
Why? 

SUG.  3.  By  what  arc  is  the  Z.  B  A  M  measured  ? 
Why? 

SUG.  4.  Then,  by  what  arc  is  the  Z  B  A  C  measured  ? 

Therefore 

PROPOSITION  XXII.    THEOREM. 

179,  The  angle  formed  by  two  secants,  meeting 
without  the  circle,  is  measured  by  one  half  the  differ- 
ence of  the  intercepted  arcs. 

.A 


L,et  A  B  and  A  C  repi'esent  two  secants,  meeting  at  A9 
without  the  circle,  and  forming  trie  angle  SAC. 

To  prove  that  the  angle  A  is  measured  by  one  half  the 
difference  of  the  arcs  B  C  and  E  D. 

SUG.  1.  Through  E,  draw  a  chord  E  M\\  to  A  B. 

SUG.  2.  By  what  arc  is  the  Z.  ME  C  measured? 
Why? 

SUG.  3.  Compare  the  Z.  A  with  the  Z  ME  C. 

SUG.  4.  Then,  by  what  arc  may  Z.  A  be  said  to  be 
measured  ? 

SUG.  5.  Express  the  arc  M  C  in  terms  of  B  C  and  D  E. 

SUG.  6.  Then,  by  what  arc  may  Z.  A  be  said  to  be 
measured  ? 

Therefore 


116  PLANE    GEOMETRY. 


PROPOSITION  XXIII.     THEOREM. 

180.  An  angle  formed  by  two  intersecting  chords  is 
measured  by  one  half  the  sum  of  the  intercepted  arcs. 


Let  A  JB  and  C  D  represent  two  chords  intersecting  at  X. 

To  prove  that  the  angle  A  X  D  is  measured  by  one  half 
the  sum  of  the  arcs  C  B  and  A  D. 

SUG.  1.  Through  D,  the  extremity  of  the  chord  CD, 
draw  D  M\\  to  A  B,  the  other  chord. 

SUG.  2.  Compare  Z   CXB   with  Z  C  D  M.      Give 
auth. 

SUG.  3.  Then,  by  what   arc   is  /.CXB  measured? 
Why? 

SUG.  4.  Express  the  arc  CM  in  terms  of  B  C  and  B  M, 
and  then  in  terms  of  B  C  and  A  D. 

SUG.  5.  Then,  by  what  arc  may  the  Z.  C  X  B  be  said 
to  be  measured. 

Therefore 

181.  A  polygon  is  said  to  be  inscribed  in  a  circle 
when  each  angle  of  the  polygon  is  an  in- 
scribed angle;  as  polygon  A  B  C  D. 

When  a  polygon  is  inscribed  in  a  circle, 
the  circle  is  said  to  be  circumscribed     \ 
about  the  polygon. 


THE   CIRCLE. 


117 


A  polygon  is  circumscribed  about  a 
circle  when  each  of  its  sides  is  tangent  to 
the  circle;  as  polygon  A  B  C  D. 

When  a  polygon  is  circumscribed  about 
a  circle,  .the  circle  is  said  to  be  inscribed 
in  the  polygon. 


PROPOSITION  XXIV.     THEOREM. 

1 82.  If  two  circles  are  tangent  to  each  other  the  line 
joining  their  centers  passes  through  the  point  of  con- 
tact* 


Let  the  circles  O  and  S  be  tangent  to  each  other,  X  the 
point  of  contact,  and  O  S  the  line  joining  ttie  centers. 

To  prove  that  O  S,  extended  if  necessary,  passes  through 
the  point  of  contact  X. 

SUG.  1.  Draw  a  common  tangent,  A  B,  to  the  two 
circles. 

SUG.  2.  At  the  point  of  contact,  X,  erect  a  J_  to  A  B, 
and  extend  it  through  both  circles. 

SUG.  3.  Why  will  the  _L  pass  through  the  centers  ? 

SUG.  4.  What  relation  will  this  line  sustain  to  O  S  ? 
Give  auth. 

Therefore 


118  PLANE  GEOMETRY. 

PROBLEMS  OF  CONSTRUCTION. 

183.  In  the   previous   work   the   constructions  have 
been  represented,  instead  of  being  actually  performed. 
When  it  is  known  that  a  certain  relation  of  points,  lines, 
or  surfaces  is  possible,  it  has  been  considered  sufficient 
to  represent  that  relation  rather  than  to  actually  con- 
struct it. 

Since  elementary  geometry  deals  only  with  figures 
which  can  be  made  from  straight  lines  and  circumfer- 
ences of  circles,  the  constructions  of  elementary  geometry 
are  those  which  it  is  possible  to  effect  by  means  of  a 
straight  edge  and  dividers,  which  are  necessitated  by  the 
postulates  named  below. 

Problems  of  construction  belong  in  no  sense  to  pure 
geometry, .but  are  simply  applications  of  the  principles 
demonstrated  in  pure  geometry. 

POSTULATES  OF  CONSTRUCTION. 

184.  Let  it  be  granted: 

1.  That  a  straight  line  may  be  drawn  between  any  two 
points  and  may  be  extended  to  any  length  through  either 
extremity. 

2.  That  a  circle  may  be  drawn  with  any  point  as  a 
center,  and  with  any  straight  line  as  a  radius. 

Note. —  It  will  be  observed  that  the  first  postulate  necessitates  the 
straight  edge,  and  the  second  the  dividers. 


Ex.  108.  If  the  middle  points  of  the  three  sides  of  a 
triangle  be  joined  by  straight  lines,  the  triangle  is  divided 
into  four  triangles  which  are  equal  in  all  respects. 

Ex.  109.  A  diameter  of  a  circle  is  greater  than  any 
other  chord. 


THE   CIRCLE.  119 


PROPOSITION  XXV.     PROBLEM. 
185.  To  bisect  a  given  straight  line. 


Let  ABbe  given^a  straight  line. 

To  bisect  A  B. 

SUG.  1.  Use  the^truth  that  if  two  equal  circles  inter- 
sect, the  line  joining  their  centers  is  bisected  at  right 
angles  by  their  common  chord. 

SUG.  2.  To  use  this  truth,  A  and  B  must  be  the  cen- 
ters of  two  equal  intersecting  circles.  Hence,  construct 
two  equal  intersecting  circles,  with  A  and  B  as  centers, 
by  postulate  2,  and  draw  their  common  chord  by  postu- 
late 1. 

Note. — A  little  experience  will  suggest  how  to  omit  the  unessential 
parts  of  lines.  For  instance,  in  the  above  construction  all  the  cir- 
cumferences may  be  omitted,  except  short  arcs  near  the  points  of  in- 
tersection, M  and  N. 


Ex.  110.  The  sum  of  the  bases 

of  a  trapezoid  is  equal  to  twice  the       \"~~7 \ 

diameter  joining  the  middle  point       \  /  \ 

d\Z- A 

of  the  legs. 

Prove  A  B  equals  one  half  the 


sum  of  C  D  and  E  G.  EN 

SUG.  Draw  M  N  ||  to  D  G,  through  A. 


120  PLANE    GEOMETRY. 

PROPOSITION  XXVI.     PROBLEM. 

186.  To  erect  a  perpendicular  to  a  given  line,  at  a 
given  point  in  that  line. 


'T-. 

A 


Let  A  B  be  the  given  line,  and  M  the  given  point. 

To  erect  a  perpendicular  to  A  B,  at~M. 

SUG.  Use  the  same  principle  as  in  the  preceding  prob- 
lem. 

In  order  to  use  the  principle,  M  must  be  the  middle 
point  of  the  line  joining  the  centers  of  the  Os.  Hence, 
find  points  on  the  line  A  B  equally  distant  from  M  (post- 
ulate 2),  and  proceed  as  in  the  preceding  problem. 

SECOND  METHOD. 

SUG.  1.  Use  the  two  principles  that  only  one  straight 
line  can  be  drawn  through  two  points  (Ax.  11),  and 
that  all  the  points  in  the  _L,  at  the  middle  point  of  a 
straight  line,  are  equally  distant  from  the  extremities  of 
the  line.  Hence,  if  two  points  of  the  _L  can  be  found, 
the  line  drawn  through  them  must  be  the  J_. 

SUG.  2.  What  point  of  the  J_  is  given  and  may  be 
used  as  one  of  the  two  points  ? 

SUG.  3.  To  find  another  point  see  Prop.  XXV. 


THE    CIRCLE.  121 


THIRD  METHOD. 

SUG.  1.  Use  the  truth  that  an  /_  inscribed  in  a  semi- 
circle is  a  rt.  <</. 

SUG.  2.  Take  any  point  O,  not  in  the  line  A  B,  as  a 
center,  and  with  a  radius  equal  to  O  M  describe  a  O.  It 
will  cut  A  B  in  M,  and  some  other  point  N.  Draw  a 
diameter  through  N.  Complete  the  solution. 

PROPOSITION  XXVII.     PROBLEM. 

187.  To  draw  a  perpendicular  to  a  given  line  from 
a  given  point  without  that  line. 


Let  A  B  represent  the  given  line,  and  Mtlie  given  paint. 

To  draw  a  perpendicular  from  M  to  the  line  A  B, 

SUG.  1.  As  Afis  one  point  in  the  required  J_,  but  one 
other  point  has  to  be  determined  in  order  to  construct  it. 

SUG.  2.  Use  the  truth  that  equal  oblique  lines  drawn 
from  a  point  in  a  _L  to  a  given  line,  cut  off  equal  dis- 
tances from  the  foot  of  the  J_. 

How  can  these  points  at  equal  distances  from  the  foot 
of  the  JL  be  found  ?  Give  auth. 

SUG.  3.  Then,  find  the  foot  of  the  _L     (Prop.  XXV.) 


122  PLANE    GEOMETRY. 

SECOND  METHOD. 

x" 
x"" 


c 

Let  A  B  represent  the  given  line,  and  M  the  given  point. 

SUG.  1.  Connect  M  with  some  point  of  the  line  A  B\ 
as  C 

SUG.  2.  Can  a  O  be  described  upon  C  M,  as  a  diam- 
eter ?  How  ?  What  problem  and  postulate  are  used  in 
this  construction  ? 

SUG.  3.  Complete  the  construction.  Give  authority 
for  each  step  in  the  solution. 

PROPOSITION  XXVIII.     PROBLEM. 
188.  To  bisect  a  given  arc. 


O. 


Let  A  1?  represent  tJie  given  arc* 

To  bisect  the  arc  A  B. 

SUG.  1.  What    proposition    has    beeii    demonstrated 
which  involves  the  bisection  of  an  arc  ?     Art.  149. 

SuG.  2.  Solve  the  problem  by  using  O,  the  center  of 
the  circle.  Also,  solve  without  using  the  center  O. 
Art.  150. 

Note. —  Do  not  use  the  references  giveu  if  you  can  answer  the  sug- 
gestive questions  without  their  aid. 


THE     CIRCLE.  123 


PROPOSITION  XXIX.     PROBLEM. 
189.  To  bisect  a  given  angle. 

L 


Let  O  represent  a  given  angle. 

To  bisect  the  angle  O. 

SUG.  1.  With  O  as  a  center  describe  an  arc. 
SUG.  2.  What  truth  has  been  demonstrated  concern- 
ing the  bisection  of  an  angle  ?     See  Arts.  171  and  188. 

PROPOSITION  XXX.     PROBLEM. 

190.  To  -find  tine  center  of  a  circle,  when  any  arc  of 
the  circumference  is  given. 
C 


Let  A  B  represent  an  arc. 

To  find  the  center  of  the  circle  of  which  A  C  B  is  an  arc. 

SUG.  1.  Select  any  three  points  on  the  arc;  as  A,  C 
and  B. 

SUG.  2.  What  is  the  locus  of  points  equally  distant 
from  A  and  C? 

SUG.  3.  What  is  the  locus  of  points  equally  distant 
from  C  and  B  ? 

SUG.  4.  Will  these  two  loci  intersect  ?     Why  ? 

SuG.  5.  Is  there  a  point  equally  distant  from  the  three 
points  A,  C  and  B  ?  Is  there  more  than  one  such  point  ? 
Complete  the  solution. 


PLANE    GEOMETRY. 


PROPOSITION  XXXI.     PROBLEM. 

191.  At  a  given  point  in  a  given  line,  to  construct 
an  angle  equal  to  a  given  angle,  with  the  given  line 
as  one  side. 


Let  M represent  the  given  angle,  A  "B  the  giten  line  and 
A  tlie  given  point. 

To  construct  at  A,  an  angle  equal  to  the  angle  M,  hav- 
ing A  B  for  one  side. 

SUG.  1.  Use  the  truth  that,  in  equal  Os,  equal  chords 
subtend  equal  angles  at  the  center. 

SUG.  2.  To  apply  this  truth,  construct  an  arc  and  its 
chord  subtending  the  /_  M\  then,  with  A  as  a  center  and 
with  the  same  radius,  construct  an  arc  intersecting  A  B. 

SUG.  3.  From  the  point  of  intersection,  construct  a 
chord  equal  to  the  chord  sub  tending*  the  ^  M. 

Name  the  postulates  used  at  each  step. 


Ex.  111.  If  fromtvvo 
opposite  vertices  of  a 
parallelogram  two  lines 
be  drawn  to  the  middle 
points  of  two  opposite 
sides,  the  lines  will  tri- 
sect the  diagonal  joining  the  other  vertices.  See  exer- 
cise 89. 


-   THE  CIRCLE.  125 

PROPOSITION  XXXII.     PROBLEM. 

192.   Through  a  given  point  to  construct  a  straight 
line  parallel  to  a  given  straight  line. 


B 


Let  A  B  be  a  given  straight  line,  and  M  a  given  point. 

To  construct  a  straight  line  through  M  parallel  to  A  B. 

SUG.  1.  Connect  M  with  any  point  of  A  B\  as  the 
point  O. 

SUG.  2.  What  relation  must  the  Z.  formed  by  the  re- 
quired line  and  M  O  bear  to  the  Z.  A  O  Ml 

Complete  the  solution. 

SECOND  METHOD. 

SUG.  1.  How  must  the  distances  between  A  B  and  the 
required  line  compare  ? 

SuG.  2.   From  M,  draw  a  _L  to  A  B. 
SUG.  3.  Draw  any  other  _L  to  A  B. 
Complete  the  solution. 


Ex.  112.  If  the  middle  points  of  two 
opposite  sides  of  a  quadrilateral  be 
joined  to  the  middle  points  of  the  diag- 
onals, the  joining  lines  form  a  parallelo- 
gram. 


126  PLANE    GEOMETRY. 

PROPOSITION  XXXIII.     PROBLEM. 

193.  To  construct  (1)  the  complement  of  a  given 
angle;  (2)  the  sivpplement  of  a  given  angle;  (3)  the 
third  angle  of  a  triangle,  having  given  the  other  two 
angles. 

SUG.  1.  In  (1)  use  the  truth  that  the  acute  Zs  of  a 
rt.  A  are  complements  of  each  other. 

SUG.  2.  In  (2)  use  the  truth  that  the  sum  of  the  Zs 
on  the  same  side  of  a  straight  line  equals  two  rt.  ^/s. 

SUG.  3.  Construct  the  two  given  ^s  so  that  they  will 
be  adjacent,  and  then  use  the  same  truth  as  in  Sug.  2. 

PROPOSITION  XXXIV.     PROBLEM. 

194.  Given  two  angles  and  the  included  side  of  a 
triangle,  to  construct  the  triangle. 


D 


Let  A  and  J5  represent  the  two  given  angles,  and  C  D 
the  included  side. 

To  construct  a  triangle  having  A  and  B  for  two  of  its 
angles  >  and  the  side  included  between  these  angles  equal  to 
the  line  C  D. 

SUG.  Apply  Prop.  XXXI,  so  as  to  make  C  and  D  each 
a  vertex  of  the  required  A. 


Ex.  113.  Prove  that  the  line  which  bisects  a  chord 
and  its  subtended  arc  passes,  if  extended,  through  the 
center  of  the  circle. 


THE    CIRCLE.  127 


PROPOSITION  XXXV.     PROBLEM. 

195.   Given  two  sides  and  ihe  included  angle  of  a 
triangle,  to  construct  the  triangle. 


M 


Let  M  and  N  represent  the  two  sides,  and  O  the  included 
angle. 

To  construct  a  triangle  having  M  and  N  for  iwo  of  its 
sides,  and  the  angle  foimed  by  those  sides  equal  to  the 
angle  O. 

This  problem  is  so  simple  that  the  student  can  solve  it 
without  the  aid  of  any  suggestions. 


Ex.  114.  Parallel  strright  lines,  included  between  par- 
allel straight  lines,  are  equal. 

Ex.  115.  What  is  the  locus  of  the  middle  points  of  all 
the  chords  of  a  circle  which  are  parallel  to  a  given  line. 

Ex.  116.  A  radius  drawn  to  the  middle  point  of  an  arc 
will  bisect  its  subtended  chord. 

Ex.  117.  A  straight  line  cannot  cut  the  circumference 
of  a  circle  in  more  than  two  points. 

SUG.  If  it  cut  in  three  points,  how  many  equal  lines 
could  there  be  drawn  from  a  point  to  a  straight  line  ? 
How  many  straight  lines  can  be  drawn  from  a  point  to  a 
straight  line  ? 


128  PLANE    GEOMETRY. 

PROPOSITION  XXXVI.     PROBLEM. 

1 96.  To  construct  a  triangle  whose  sides  are  three 
given  lines. 

A B 

r. D 


Let  A  B,  C  D  and  E  F  represent  three  given  lines. 

To  construct  a  triangle  whose  sides  are  A  B>  C  D  and 
E  F  respectively. 

SUG.  1.  Draw  the  line  A  B.     How  many  vertices  of 
the  required  A  are  now  located  ? 

SUG.  2.  With  the  extremity  C,  of  the  line  C  D,  fixed 
at  A)  what  is  the  locus  of  the  point  D? 

SUG.  3.   From  the  answer  to  Sug.  2,  where  does  the 
third  vertex  of  the  required  A  lie  ? 

SUG.  4.  With  the  extremity  E,  of  the  line  E  F,  fixed 
at  B,  what  is  the  locus  of  the  point  F1 

SUG.  5.  From  the  answer  to  Sug.  4,  where  does  the 
third  vertex  of  the  required  A  lie  ? 

,   SUG.  6.  Now,  give  complete  directions  for  construct- 
ing the  A. 

Can  three  such  lines  be  given  that  the  problem  is  im- 
possible ? 


Ex.  118.  Prove  that  the  two  tangents 
from  a  point  to  a  circle  are  equal. 

SUG.  Connect  the  center  of  the  circle 
with  the  given  point.  Complete  the 
construction. 


THE    CIRCLE. 


129 


EXERCISES. 

119.  If  a  tangent  and  a  chord  of  a  circle  are  parallel, 
prove  that  the  arcs  intercepted  between  the  tangent  and 
chord  are  equal. 

120.  Upon  a  given  base,  construct  an  isosceles  triangle, 
in  which  the  sum  of  two  equal  sides  shall  equal  a  given 
line. 

121.  All  angles  inscribed  in  the   same   segment  are 
equal. 

122.  Prove  that  an  angle  inscribad  in 
a  semicircle  is  a  right  angle. 

To  prove  that  the  angle  B  A  C  is  a 
right  angle. 

123.  Demonstrate  Prop.  XXII  by  an- 
other method. 

Sue.  Draw  B  E.     By  what   arc 


s 


D  E  meas- 
Then, by 


ured  ?     By  what  arc  is  Z.  B  E  C  measured  ? 
what  arc  must  Z.  A  be  measured  ? 

124.  In  the  same   circle,  or  in   equal 
circles,  an  angle  inscribed  in  the  smaller  ^ 
of  two  segments  is  larger  than  an  angle  ^ 
inscribed  in  the  larger  segment. 

Prove  that  the  angle  A  E  B  is  larger 
than  the  angle  COD. 

125.  An  angle  formed  by  a  tangent  and  a  secant  is 
measured  by  one  half  the  difference  of  the  intercepted 
arcs. 

126.  The  segments  of  a  straight  line 
intercepted    by    concentric    circles     are 
equal. 

Prove  A  B  equals  CD. 
Concentric  circles   are  circles  having 
the  same  center. 
9-Geo. 


130  PLANE   GEOMETRY. 

PROPOSITION  XXXVII.     PROBLEM. 

197.   Through  a  given  point,  to  draw  a  tangent  to  a 
given  circle. 

CASE  I.      When  the  point  is  on  the  circumference. 


Let  A  represent  the  given  paint  in  tlie  circumference  of 
the  given  circle. 

To  draw  a  tangent  to  the  circle  O,  through  the  point  A. 

SUG.  1.  What  relation  does  a  tangent  bear  to  the  ra- 
dius drawn  to  the  point  of  contact  ? 

SUG.  2.  Give  the  method  of  making  the  construction 
in  this  case. 

II.      When  the  given  point  is  without  the  circle. 

A 


Let  A  represent  the  given  point  without  the  circle  O. 

To  draw  a  tangent  to  the  circle  O,  through  the  point  A. 


THE   CIRCLE.  131 


SUG.  1.  Connect  O  and  A. 

SUG.  2.  If  the  required  point  of  contact  were  joined 
both  to  O  and  A,  what  kind  of  an  Z.  would  be  formed  ? 

SUG.  3.  What  is  the  locus  of  the  vertex  of  the  Z  in  a 
A  whose  base  is  O  A  ?  See  Art.  177,  and  Bxs.  121, 122. 

SUG.  4.  Give,  now,  complete  directions  for  finding  the 
point  of  contact,  and  hence  for  constructing  the  tangent. 

PROPOSITION  XXXVIII.     PROBLEM. 

198.  To  circumscribe  a  circle  about  a  given  tri- 
angle. 


Let  ABC  represent  the  given  triangle. 

To  circumscribe  a  circle  about  the  triangle  ABC. 

SUG.  1.  The  problem  is  to  find  the  center  of  a  O  whose 
circumference  passes  through  A,  B  and  C;  i.  e.,  to  find 
a  point  equally  distant  from  A,  B  and  C. 

SUG.  2.  See  Art.  158,  and  then  give  complete  direc- 
tions for  circumscribing  a  O  about  the  A  A  B  C. 

QUERY.  How  many  circles  can  be  circumscribed  about 
a  triangle  ?  Why  ? 


132  PLANE    GEOMETRY. 

PROPOSITION  XXXIX.     PROBLEM. 
199.  To  inscribe  a  circle  in  a  given  triangle* 


Let  ABC  represent  the  triangle. 

To  inscribe  a  circle  in  the  triangle  ABC. 

SUG.  1.  If  a  O  can  be  inscribed  in  the  A,  the  center 
of  the  O  must  be  equally  distant  from  the  three  sides. 

SUG.  2.  Is  there  such  a  point?    See  exercise  71. 

SUG.  3.  How  may  this  point  be  found  ? 

SUG.  4.  Give  complete  directions  for  inscribing  a  circle 
in  a  given  A. 

QUERY.  How  many  circles  can  be  inscribed  in  a  given 
triangle?  Why? 


PROPOSITIONS  IN  CHAPTER  II. 


PROPOSITION  I. 

Two  circles  are  equal  if  the  radius  of  one  equals  the  radius  of  the 
other. 

PROPOSITION  II. 

A  diameter  divides  a  circle  into  two  equal  parts. 

PROPOSITION  III. 

In  the  same  circle,  or  in  equal  circles,  equal  angles  at  the  center 
intercept  equal  arcs  at  the  circumference. 

PROPOSITION  IV. 

CONVERSE  OF  PROP.  III.  In  the  same  circle,  or  in  equal  circles, 
equal  arcs  subtend  equal  angles  at  the  center. 

PROPOSITION  V. 

In  the  same  circle,  or  in  equal  circles,  chords  which  subtend  equal 
arcs  are  equal. 

PROPOSITION  VI. 

CONVERSE  OF  PROP.  V.  In  the  same  circle,  or  in  equal  circles,  arcs 
which  are  subtended  by  equal  chords  are  equal. 

PROPOSITION  VII. 

In  the  same  circle,  or  in  equal  circles,  two  chords  which  subtend 
unequal  arcs  are  unequal,  that  chord  being  greater  which  subtends 
the  greater  arc. 

PROPOSITION  VIII. 

CONVERSE  OF  PROP.  VII.  In  the  same  circle,  or  in  equal  circles, 
two  arcs  which  are  subtended  by  unequal  chords  are  unequal,  that 
arc  being  greater  which  is  subtended  by  the  greater  chord. 

PROPOSITION  IX. 

A  radius  which  is  perpendicular  to  a  chord  bisects  the  chord  and 
its  subtended  arc. 


134  PLANE   GEOMETRY. 


PROPOSITION  X. 

A  line  perpendicular  to  a  chord  at  its  middle  point  passes  through 
the  center. 

PROPOSITION  XI. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  are  equally 
distant  from  the  center,  and  of  two  unequal  chords  the  greater  is 
nearer  the  center. 

PROPOSITION  XII. 

A  straight  line  perpendicular  to  the  radius  of  a  circle  at  its  extrem- 
ity is  tangent  to  the  circle. 

PROPOSITION  XIII. 

CONVERSE  OF  PROP.  XII.  If  a  straight  line  is  tangent  to  a  circle 
the  radius  meeting  it  at  the  point  of  contact  is  perpendicular  to  it. 

PROPOSITION  XIV. 
Arcs  of  a  circle  intercepted  by  parallel  chords  are  equal. 

PROPOSITION  XV. 

Through  three  points,  not  in  the  same  straight  line,  one  circumfer- 
ence, and  only  one,  is  possible. 

PROPOSITION  XVI. 
To  find  the  greatest  common  unit  of  measure  of  two  given  lines. 

PROPOSITION  XVII. 

Given  the  greatest  common  unit  of  measure  of  two  quantities  to 
find  their  ratio. 

PROPOSITION  XVIII. 

If  two  variables  are  always  equal  as  they  approach  their  limits, 
their  limits  are  equal. 

PROPOSITION  XIX. 

In  the  same  circle,  or  in  equal  circles,  two  angles  at  the  center 
have  the  same  ratio  as  the  arcs  which  they  intercept  at  the  circumfer- 
ence. 


THE   CIRCLE.  135 


PROPOSITION  XX. 
An  inscribed  angle  is  measured  by  one  half  its  intercepted  arc. 

PROPOSITION  XXI. 

An  angle  formed  by  a  tangent  and  a  chord  is  measured  by  one  half 
the  intercepted  arc. 

PROPOSITION  XXII. 

The  angle  formed  by  two  secants,  meeting  without  the  circle,  is 
measured  by  one  half  the  difference  of  the  intercepted  arcs. 

PROPOSITION  XXIII. 

An  angle  formed  by  two  intersecting  chords  is  measured  by  one 
half  the  sum  of  the  intercepted  arcs. 

PROPOSITION  XXIV. 

If  two  circles  are  tangent  to  each  other  the  line  joining  their  cen- 
ters passes  through  the  point  of  contact. 

PROPOSITION  XXV. 
To  bisect  a  given  straight  line. 

PROPOSITION  XXVI. 

To  erect  a  perpendicular  to  a  given  line,  at  a  given  point  in  that 
line. 

PROPOSITION  XXVII. 

To  draw  a  perpendicular  to  a  given  line  from  a  given  point  with- 
out that  line. 

PROPOSITION  XXVIII. 
To  bisect  a  given  arc. 

PROPOSITION  XXIX. 
To  bisect  a  given  angle. 

PROPOSITION  XXX. 

To  find  the  center  of  a  circle,  when  any  arc  of  the  circumference  is 
given. 


136  PLANE   GEOMETRY. 

PROPOSITION  XXXI. 

At  a  given  point  in  a  given  line,  to  construct  an  angle  equal  to  a 
given  angle,  with  the  given  line  as  one  side. 

PROPOSITION  XXXII. 

Through  a  given  point,  to  construct  a  straight  line  parallel  to  a 
given  straight  line. 

PROPOSITION  XXXIII. 

To  construct  (1)  the  complement  of  a  given  angle;  (2)  the  supple- 
ment of  a  given  angle;  (3)  the  third  angle  of  a  triangle,  having  given 
the  other  two  angles. 

PROPOSITION  XXXIV. 

Given  two  angles  and  the  included  side  of  a  triangle,  to  construct 
the  triangle. 

PROPOSITION  XXXV. 

Given  two  sides  and  the  included  angle  of  a  triangle,  to  construct 
the  triangle. 

PROPOSITION  XXXVI. 

To  construct  a  triangle  whose  sides  are  three  given  lines. 

PROPOSITION  XXXVII. 
Through  a  given  point,  to  draw  a  tangent  to  a  given  circle. 

PROPOSITION  XXXVIII. 
To  circumscribe  a  circle  about  a  given  triangle. 

PROPOSITION  XXXIX. 
To  inscribe  a  circle  in  a  given  triangle 


CHAPTER  III. 
PROPORTIONAL  LINES,  AND  SIMILAR  POLYGONS. 


THE  THEORY  OF  PROPORTION. 

200.  Proportion  is  an  equality  between  two  ratios. 
If  A  and  B  form  one  ratio,  and  C  and  D  an  equal  ratio, 
a  proportion  is  formed  which  may  be  written  in  either  of 
the  forms: 


Either  of  these  is  read:  "the  ratio  of  A  to  B  equals 
the  ratio  of  C  to  D  ;  "  or,  "  A  is  to  B  as  C  is  to  Z>." 

201.  The  terms  of  a  proportion  are  the  four  numbers, 
or  quantities,  compared. 

In  any  ratio  the  first  term  is  called  the  antecedent, 
and  the  second  term  the  consequent.  Hence,  in  any 
proportion  the  first  and  third  terms  are  called  anteced- 
ents, and  the  second  and  fourth  terms  are  called  conse- 
quents. 

In  any  proportion,  the  first  and  fourth  terms  are  called 
extremes,  and  the  second  and  third  terms  are  called 
means. 

A          C 
In  the  proportion  -  -  =  — ,  the  first  term  is  A,  the 

£j  JLS 

second  B,  the  third  C,  and  the  fourth  D.  A  and  C  are 
the  antecedents,  B  and  D  are  the  consequents,  A  and  D 
are  the  extremes,  and  B  and  C  are  the  means. 


138  PLANE    GEOMETRY. 

PROPOSITION  I.    THEOREM. 

201.  In  a  proportion,  all  of  whose  terms  are  num- 
bers, the  product  of  the  means  equals  the  product  of 
the  extremes. 

Let  -^  =  -=-   be  a  proportion  in  tvhich  A,  B,  C  and  D 

Jo  JLJ 

are  numbers. 

To  prove  A  D  =  B  C. 

SUG.  1.  What  process  is  performed  upon  the  expression 
A 
TT  to  produce  the  product  ADI 

£> 

Q 

SUG.  2.  What,    then,    must    the    expression    -=:     be 

changed  into  ?     Why  ? 
Therefore 

MODEL. 

PROPOSITION  I.     THEOREM. 

202.  In  a  proportion,  all  of  whose  terms  are  num- 
bers, the  product  of  the  means  equals  the  product  of 
the  extremes. 

Let  -^    =  -=r   be  a  proportion  in  which  A,  B,  C  and  D 
j>         u 

are  numbers. 

To  prove  A  D  =  B  C. 

~  X  B  D  =  A  D  and  ~    X   B  D  =  B  C. 

Hence,  by  Ax.  4,  A  D  =  B  C. 

Therefore,  in  a  proportion,  all  of  whose  terms  are  num- 
bers, the  product  of  the  means  equals  the  product  of  the 
extremes 


PROPORTION  —  SIMILAR   POLYGONS.  139 

PROPOSITION  II.    THEOREM. 

203.  If  the  product  of  two  numbers  equals  the 
product  of  two  other  nuiribers,  the  factors  of  one 
product  may  be  made  the  means,  and  the  factors  of 
the  other  product  the  extremes,  of  a  proportion. 

Let  AB=  C  D  be  an  equation,  in  which  A,  B,  C  and  D 
are  tlie  numbers. 

To  prove  £-§. 

SUG.  1.  What  process  must  be  performed  upon  A  B  to 

^ 
produce  the  fraction  ^_-  ? 

SUG.  2.  By  the  same  process  the  second  member,  C  D, 
is  changed  into  what? 
Therefore 

Note. —  It  frequently  occurs  that  expressions  are  used  which  are 
perfectly  intelligible  in  some  cases  and  have  no  meaning  in  other  cases. 
For  example,  the  expression  A  B  (that  is,  A  times  B}  has  a  perfectly 
definite  meaning,  if  A  and  B  are  both  numbers;  it  also  has  a  definite 
meaning  when  A  is  a  pure  number  and  B  a  quantity,  provided  A  be  con- 
sidered as  the  multiplier;  but  if  A  and  B  are  both  quantities,  A  B  has 

no  meaning.     Likewise,  the  expression    -=    has   a  definite   meaning 

if  A  and  B  are  both  numbers;  it  also  has  a  definite  meaning  if  A  is  a 
quantity  and  B  a  number;  also,  if  A  and  B  are  both  quantities  of  the 
same  kind;  but  has  no  meaning  if  A  is  a  number  and  B  is  a  quantity, 
or  if  A  and  B  are  quantities  unlike  in  kind.  See  Art.  161. 

In  all  the  following  operations  upon  ratio,  and  applications  of  it,  care 
should  be  taken  to  interpret  the  symbols  of  number  or  quantity,  and 
see  if  the  principles  of  the  fundamental  rules  of  arithmetic  and  al- 
gebra can  be  applied.  The  scholia  following  the  theorems  in  the  the- 
ory of  proportion  will  suggest  the  limitations  which  the  principles  un- 
derlying the  operations  impose  upon  the  propositions. 


140  PLANE   GEOMETRY. 

204.  A  mean  proportional  between  two  given  num- 
bers, or  quantities,  is  the  second  or  third  term  of  a  pro- 
portion in  which  the  means  are  alike,  and  the  extremes 
are  the  given  numbers  or  quantities.     In  the  proportion 

A.         B 

-n   =  -f?  i  B  is  a  mean  proportional  between  A  and  C. 

PROPOSITION  III.     PROBLEM. 

205.  To  determine  a  mean  proportional  between 
two  given  numbers. 

Liet  A  and  C  be  the  given  nutribers,  and  B  the  required 
mean  proportional, 

To  determine  B. 

SUG.  In  the  equation   -=-  =  -=  find  the  value  of  B. 
13          C 

Note. —  In  propositions  I,  II  and  III,  the  terms  of  the  propor- 
tions must  be  regarded  not  as  geometric  magnitudes,  but  as  numbers 
or  the  measures  of  magnitudes.  These  propositions,  however,  afford 
aid  in  investigating  propositions  whose  terms  are  geometric  magni- 
tudes instead  of  numbers.  For  example,  suppose  it  is  required  to 
determine  the  mean  proportional  between  two  lines,  A  and  C,  in  which 
a  rod  is  taken  as  the  unit  of  measure.  If  B  is  the  required  mean  pro- 

A  /? 

portional,  the  equation  -=   =  -=  must   hold.     Since  A   and    C   are 

each  some  number  of  rods,  B  must  also  be  some  number  of  rods. 
Suppose  A  —  a  rods, 

B  =  b  rods, 
C  =  c  rods. 

Now,  B  is  determined  as  soon  as  b  is  determined.     From  the  pro- 
portion -=  =  —   it  follows  that  —  =   —  is  true,  and  from  this  last 

jD  C  O  C 

equation  b  may  be  found. 


Ex.  127.  An  angle  formed  by  two  tangents  is  meas- 
ured by  one  half  the  difference  of  the  intercepted  arcs. 


PROPORTION  —  SIMILAR    POLYGONS.  141 

PROPOSITION  IV.    THHOREM. 

206.  If,  in  several  successive  ratios,  the  consequent 
of  the  first  equals  the  antecedent  of  the  second,  the  con- 
sequent of  the  second  equals  the  antecedent  of  the 
third,  etc. ,  the  ratio  of  the  antecedent  of  the  first  to 
the  consequent  of  the  last  equals  the  product  of  tlie 
ratios. 

Let  -g-,    ^-,    -=j,   -=    and  -^    represent  the  ratios. 

To  prove  that  -^  equals  the  product  of  the  ratios. 

SUG.  1.  If  A,  B,  C,  D,  E  and  F  are  numbers,  apply 
the  algebraic  rule  for  multiplying  fractions. 

SUG.  2.  If  A,  B,  C,  D,  E  and  F  are  geometric  magni- 
tudes, they  are  all  magnitudes  of  the  same  kind,  and 
by  definition  each  ratio  is  a  number. 

^  =  w     £  ^  w      C_        r     D_  =         <£.«,. 

m,  n,  r,  s  and  /  b2ing  numbers. 

SUG.  3.  Then,  A  =  m  B,  B  =  n  C,  C  =  r  D, 
D  =  s  E  and  E  =  /  F.  Why  ? 

SUG.  4.  Since  A  —  m  B  and  B  —  n  C,    A  =  m  n  C. 

Since  A  =  m  n  C  and  C  =  r  D,    A  =  m  n  r  D. 

In  the  same  way  A  =  m  n  r  s  E. 

Finally,  A  =  m  n  r  s  t  F. 

SUG.  5.  Since  A^mnrstF,    -=  —mnrst.    Why> 

r 

SUG.  6.  Notice,  now,  that  m  n  r  1 1  is  the  product  of 
the  given  ratios. 
Therefore 


142  PLANE    GEOMETRY. 


Note  j. —  A  careful  interpretation  of  the  symbols  should  be  made 
at  each  step  of  the  preceding  proposition. 

Note  2. —  The  preceding  proposition   may  be  illustrated  as  follows: 

1  bushel    _  1  peck     _          1  quart    _ 

1  peck          "     '      1  quart   ""    '      1  pint 

Therefore,  - — US          =  64.  the  product  of  the  ratios. 
1  pint 

To  say  that —  =  4,  is  only  another  form  of  saying  that 

1  bushel  —  4  pecks.     -=  =  m  is  only  another  form  of  A  =  m  B. 

This  should  be  illustrated  until  the  pupil  sees  that  they  are  only 
different  forms  of  expressing  the  same  thought,  viz.,  that  1  peck  is 
the  unit  by  which  1  bushel  is  measured,  or  that  B  is  the  unit  by 
which  A  is  measured,  in  whichever  way  it  is  expressed. 

QUERY.  In  the  equation  A  —  m  n  r  D,  which  sym- 
bols represent  numbers  and  which  represent  quantities  ? 

PROPOSITION  V.     THEOREM. 

207.  Both  terms  of  a  ratio  may  be  multiplied  by 
any  number  without  changing  the  value  of  the  ratio. 

Let  T—  represent  a  ratio. 

~  A         m  A  . 

To  prove  -=?  =  — ^-,  m  being  a  number. 
ti         m  B 

A 

SUG.  1.  I^et  the  ratio  -^  equal  the  number  r. 
r> 

^ 

SUG.  2.   In  the  equation  -=-  =  r,  find  what  A  equals. 

B 

SUG.  3.   Multiply  both  members  of  the  equation  last 
found  by  m}  and  from  the  result  find  the  value  of  the  ra- 
m  A 


tio 


m  B  ' 

SUG.  4.  How,  then,  does  -jc  compare  with   — ~, 
Therefore 


PROPORTION  —  SIMILAR   POLYGONS.  14.3 

208.  COROLLARY.     Both   terms   of   a   ratio   may   be 
divided  by  the  same  number;  also,  by  the  same  quantity, 
provided  it  [is  a  quantity  of  the  same  kind  as  the  terms 
of  the  given  ratio. 

209.  SCHOLIUM.     In  the  foregoing  proposition  A  and 
B  may  both  be  numbers,  or  both  like  quantities. 

PROPOSITION  VI.    THEOREM. 

210.  If  four  numbers,  or  like  quantities,  are  in  pro- 
portion, the  ratio  of  the  first  to  the  third  equals  the 
ratio  of  the  second  to  the  fourth. 

.   A       C 


To  prove  ^    =  —  - 

A  C 

SUG.  1.  If  2j-   =  the  number  m,  what  does  -=  equal? 

SUG.  2.  Find  A  and  C  in  terms  of  B  and  D,  respec- 
tively. 

Express  the  ratio  of  A  to  C,  and  reduce  by  Prop.  V, 
Cor. 

Therefore  -- 

211.  SCHOLIUM.  Proposition  VI  practically  states 
that  the  second  and  third  terms,  of  a  proportion  may  be 
interchanged.  This  is  true  if  the  terms  of  the  propor- 
tion are  numbers,  or  quantities  of  the  same  kind,  but  not 
true  if  the  terms  are  quantities  unlike  in  kind.  For  ex- 

.      .,  .        16  rods         1  peck   . 

ample,  the  proportion  -^  -  j-  =  7-^  —  is   true,  since 
2  rods         1  pint 

each  ratio  equals  8,  but  if  the  second  and  third  terms  be 

interchanged  the  result  would  be    -=  --  —  =  -  —  :  —  ,  an 

1  peck        1  pint 

absolutely  meaningless  expression. 


PLANE   GEOMETRY. 


212.  When  from  any  proportion  a  new  proportion  is 
obtained  by  taking  the  antecedents  for  one  ratio,  and  the 
consequents,  in  the  same  order,  for  the  other  ratio,  the 
second  proportion  is  said  to  be  deduced  from  the  first  by 
alternation. 

MR...  .       .         M       N 

--  =  —  is  deduced  by  alternation  from  -5-  =  -^-. 

YV       o  /to 

PROPOSITION  VII.    THEOREM. 

213.  If  four  numbers,  or  quantities,  are  in  propor- 
tion, the  ratio  of  the  second  to  the  first  equals  the  ra- 
tio of  the  fourth  to  the  third. 

A  (^ 

Let  ^-  =  -=r  be  the  given  proportion. 
Jt>        u 

B         D 
fo  prove  --=-£. 

A  C 

SUG.  1.  L/et  -=r  =  m.     Then  -—  =  m. 

JD  JJ 

SUG.  2.  From  the  equations  in  Sug.  1,  find  the  values 
of  A  and  C  respectively. 

SUG.  3.  Divide  by  A  and  C,  respectively,  the  two 
equations  obtained  in  Sug.  2.  Compare  the  results  and 
reduce. 

Therefore  - 

214.  SCHOiyiUM.     Proposition   VII   states  that   in   a 
proportion  the  terms  of  one  ratio  may  be  interchanged, 
provided  the  terms  of  the  other  ratio  are  interchanged 
also.     In  this  proposition  the  terms  of  one  ratio  may  be 
different  in  kind  from  the  terms  of  the  other  ratio.     For 

16  rods        1  peck 
example,  the  proportion   -^  —  —  ==  -i     •   t     1S  true>  an(i 

the  proportion  obtained  by  applying  the  theorem,  viz., 

2  rods        1  pint    .      . 
•r-z  —  T-  =  T-Z  —  =-  is  also  true. 
16  rods        1  peck 


PROPORTION  —  SIMILAR    POLYGONS.  145 

215.  When,  from  any  proportion,  a  new  proportion  is 
formed  by  inverting  the  ratios,  the  second  proportion  is 
said  to  be  deduced  ironi  the  first  by  inversion.  * 

PROPOSITION  VIII.     THEOREM. 

216.  If  four  numbers,  or  quantities,  are  in  propor- 
tion, the  ratio  of  the  first  plus  the  second,  to  the  second, 
equals  the  ratio  of  the  third  plus  the  fourth,  to  the 
fourth. 

Let  jj  =  jrj  *>e  Mie  ffiven  proportion. 

A  +  B         C+  D 
Toprove     _____  =  ______ 

SUG.  1.  Let    -=  =  m.     Then  £  =  m. 

£>  D 

SUG.  2.   From  --  =  m,  what  does  -—7- —  equal  ? 

SUG.  3.  From  -^  =  m,  what  does  — -= —  equal  ? 

Compare  answers  to  Sugs.  3  and  4. 
Therefore 

217.  SCHOLIUM.     In  proposition  VIII,  the  terms  of 
one  ratio  may  be  different  in  kind  from  the  terms  of  the 
other  ratio  and  the  proposition  still  be  true.     For  ex- 
ample the  proportion  -  =  -     *— —•  is  true  and 

6  rods  5  quarts 

the  proportion  obtained  by  applying  the  theorem,  viz., 

18  rods         15  quarts  . 
-« —     -  =»     g  -  is  also  true. 

6  rods          5  quarts 

A         C 

218.  When  from  the  proportion    -^    =  -^     the   pro- 
portion -  -  =  -  -  is   obtained,   the   second   is 

Jo  D 

said  to  be  deduced  from  the  first  by  composition. 

10— Geo. 


146  PLANE   GEOMETRY. 


PROPOSITION  IX.     THEOREM. 

219.  If  four  numbers,  or  quantities,  are  in  propor- 
tion, the  ratio  of  the  first  minus  the  second  to  the  sec- 
ond, equals  the  ratio  of  the  third  minus  the  fourth,  to 
the  fourth. 

Let  ]g  —  j)  &e  fhe  given  proportion. 

To  prove  that  -  = — —  . 

J3  JD 

A  C 

SUG.  1.   Let    -=,-  =   m.     Then  -^-  =  m. 

r>  U 

^  ^  g 

SUG.  2.   From  -=-  =  m,  what  does  -  -  equal  ? 

£J  £) 

SUG.  3.   From  -=r  =   m,  what  does    -  '—== —  equal  ? 

SUG.  4.   Compare  answers  to  Sugs.  2  and  3. 
Therefore 

220.  SCHOLIUM.     In  Proposition  IX,  the  terms  of  one 
ratio  may  be  different  in  kind  from  the  terms  of  the  other 
ratio  and  the  proposition  still  be  true.     Illustrate  this 
fact. 

A        C 

221.  When  from  the  proportion    -=r  =  -=r ,    the  pro- 

h  JJ 

•      A  —  B         C-  D    .       ,..'•- 

portion  = = =r —  is   obtained,  the  second   is 

B  L) 

said  to  be  deduced  from  the  first  by  division. 
PROPOSITION  X.     THEOREM. 

222.  In  a  series  of  equal  ratios,  all  of  whose  terms 
are  numbers  or  quantities  of  the  same  Tcind,  the  ratio 
of  the  sum  of  the  antecedents  to  the  sum  of  the  conse- 
quents equals  any  one  of  the  ratios. 


PROPORTION  —  SIMILAR   POLYGONS.  147 


B~  D  ~  F  ~  H* 

A  +  C+  E  +  G        A        C 
Toprove  B  +  D  +  F+f/  ==  ^  ==  ^  etc. 

SuG.  1.  Let  each  of  the  given  ratios  =  x. 

SuG.  2.  Put  each  ratio  separately  equal  to  x  and  find 
each  antecedent. 

SUG.  3.  Find  the  sum  of  all  the  antecedents. 

SUG.  4.  From  the  result  of  Sug.  3,  find  the  value  of  x. 

SuG.  5.  Equate  the  value  of  x  just  found  to  any  of  the 
given  ratios. 

Therefore  - 

223.  In  proposition  X,  the  terms  of  the  given  ratios 
must  be  numbers  or  quantities  of  the  same  kind.  Illus- 
trate the  absurdity  of  using  quantities  of  different  kinds. 


Ex.  128.  If  J  -  §,    Prove  that  -^  =  -^ 

A  +  B       C  +  D 

and  also  that    ^ —  =  — -= — . 

A  L, 

In  deducing  the  second  proportion  from  the  first,  are 
A,  B,  C  and  D  restricted  to  being  numbers,  or  may  they 
be  quantities  as  well,  and,  if  quantities,  are  they  all 
quantities  of  the  same  kind,  or  may  they  differ  in  kind  ? 

Ex.  129.  If  J  =  g,    prove  that  -^  =  ^-, 

A  -  B       C  -  D 

and  also  that    — — —  =  — — — . 

In  deducing  the  second  proportion  from  the  first,  are 
A,  Bt  Cand  D  restricted  in  any  way,  and,  if  so,  how  ? 


148  PLANE    GEOMETRY. 

PROPOSITION  XI.    THEOREM. 

224.  If  three  terms  of  one  proportion  are  respec- 
tively equal  to  the  three  corresponding  terms  of  an- 
other proportion,  the  fourth  terms  are  equal. 

Let±=%    an*    ±  =  %. 

To  prove  that  C  =  M. 

SUG.  1.  If  ^  =  r,    then   ~  =  r    and  ^  =  r.    Why? 

SUG.  2.  From  -=?  =  r,      find  C,      and  from     —  =  r 

find  M. 

SUG.  3.  Compare  C  and  M. 

Therefore 

Note. —  In  all  the  preceding  propositions  in  the  theory  of  propor- 
tion, the  theorems  are  true  if  the  terms  of  the  proportions  are  num- 
bers. 

If  the  terms  are  geometric  magnitudes,  the  limitations  noted  should 
be  carefully  studied  and  applied.  Read  Arts.  159,  160  and  161  on 
measurement,  and  note  their  bearing  on  the  theory  of  proportion. 


Ex.  129.  Prove  that  an  angle 
formed  by  a  tangent  and  a  chord,  is 
measured  by  one  half  the  intercepted  Afy 
arc,  using  the  following  construction: 
Drop  a  _L  from  the  center  of  the  O  to 
the  chord  and  extend  to  the  arc.  Con- 
nect O  and  the  point  of  contact  A. 

SUG  Compare  Z.  M  O  A  with  Z.  between  tangent 
and  chord. 

Ex.  130.  In  the  figure  of  Prop.  XXIII,  Chap.  II,  con- 
nect B  and  D,  and  prove  the  proposition  in  another 
manner. 


PROPORTION  —  SIMILAR   POLYGONS.  149 

PROPOSITION  XII.    THEOREM. 

225.  If  parallel  lines  intercept  equal  segments  on 
one  transversal,  they  intercept  equal  segments  on  all 
transversals. 


Go 


/ 


Let  E  F,  G  H,  etc.,  represent  parallel  lines  intercepting 
equal  segments  E  G,  G  I,  etc.,  on  the  transversal  A  B,  and 
V.w  segments  F  H,  H  M,  etc.,  on  the  transversal  C  D. 

To  prove  F  H  =  HM=  M  N,  etc. 

SUG.  1.  From  E,  G,  /,  etc.,  draw  lines  E  O,  G  Pt  IS, 
etc.,  ||  to  CD. 

SUG.  2.  In  the  As  E  O  G,  G  P /,  etc.,  compare  the 
lines  E  O,  G  P,  IS,  etc. 

SUG.  3.  Compare  E  O  with  F  H,  G  P  with  H  M,  I  S 
with  M  N,  etc. 

Therefore 


Ex.  131.  If  a  quadrilateral  is  inscribed 
in  a  circle  the  sum  of  the  opposite 
angles  equals  two  right  angles.  Prove 
that  Z  A  plus  Z.  C  equals  two  rt.  Zs. 


150  PLANE   GEOMETRY. 


132.  An  angle  formed  by  a  tangent  and  a  secant 
is  measured  by  one  half  the  difference  of  the  intercepted 
arcs. 

PROPOSITION  XIII.    THEOREM. 

226.  If  a  line  is  parallel  to  the  base  of  a  triangle, 
the  ratio  of  the  segments  of  one  side  equals  the  ratio  of 
the  segments  of  the  other  side. 


Let  DEbe  a  line  parallel  to  the  base  B  C  of  the  tri- 
angle ABC. 

AD       A  E 
To  prove  -^2=-—. 

CASE  I.      When  A  D  and  D  B  are  commensurable. 

SUG.  1.  Since  A  D  and  D  B  are  commensurable,  sup- 
pose each  to  be  measured,  and  express  their  ratio. 

SUG.  2.  Through  the  points  of  division  in  A  B,  draw 
lines  ||  to  the  base  and  intersecting  A  C. 

SUG.  3.  Compare  the  segments  of  A  £  and  E  C  with 
respect  to  size.  How  many  are  there  as  compared  with 
the  number  of  segments  in  A  B  ? 

SUG.  4.  Compare  the  ratio  -^-^    with  the  ratio   Tr^s , 

c,  L  JJ  Jo 


PROPORTION  —  SIMILAR    POLYGONS.  151 


CASE  II.      When  A  B  and  D  B  are  incommensurable. 


B 

SUG.  1.  Divide  A  D  into  any  number  of  equal  parts, 
and  lay  off  the  unit  of  measure  upon  D  B  as  many  times 
as  possible.  There  will  be  a  remainder,  MB,  less  than 
the  unit  of  measure.  Why  ? 

SUG.  2.  Through  M,  draw  M  N  ||  to  B  C 

A  D  A  F 

SUG.  3.  Compare   -prT>  with    -p-T>-     Give  auth. 

D  M  EN 

SUG.  4.  If  the  unit  of  measure  be  continually  dimin- 

A  D 
ished,  the  ratio         ^    is  a  variable.     Why  ? 

SUG.  5.  What  is  its  limit  ?     Why  ? 

A  E 
SUG.  6.  The  ratio   -^.y  is  also  a  variable.     What  is 

its  limit  ? 

SUG.  7.  How  do  the  limits  of  -^—^    and  T^>  com- 

D  M  EN 

pare  ?     (See  Art.  170.) 

Therefore  — 

Note. —  Compare  this  demonstration  with  that  of  Prop.  XIX, 
Chap.  II. 

227.  COROU,ARY.  If  a  line  be  drawn  parallel  to  the 
base  of  a  triangle,  one  side  is  to  either  of  its  segments 
as  the  other  side  is  to  its  corresponding  segment. 

SUG.  Take  the  conclusion  of  the  proposition  by  com- 
position. 


152  PLANE   GEOMETRY. 

PROPOSITION  XIV.     THEORKM. 

228.  If  a  line  divides  two  sides  of  a  triangle  pro- 
portionally, it  is  parallel  to  the  base. 


-W 

V 


Let  A  B  C  be  a  triangle,  and  let  D  E  divide  tlie  sides 

A  B  and  A  C  so  that  44^  -  4|?* 
A  Jo         AC 

To  prove  that  D  E  is  parallel  to  the  base  B  C. 

SUG.  1.  From  D,  draw  D  M  \\  to  B  C. 

A.  D  A.  M 

.SuG.  2.  Compare  the  ratios    -7-5   and         r  . 

./i  Jo  A.  C- 

SuG.  3.  See  hypothesis,  and  compare  A  R  and  A  M. 
Ax.  1. 

SUG.  4.  Complete  the  demonstration. 
Therefore 

SIMILAR  POLYGONS . 

229.  Similar  polygons  are  those  which  have  their 
corresponding  angles  equal,  and  corresponding  sides 
proportional. 

Points,  lines,  or  angles,  which  are  similarly  situated 
in  similar  polygons,  are  called  homologous. 


PROPORTION  —  SIMILAR    POLYGONS. 


153 


In  similar  polygons,  the  ratio  of  two  homologous  sides 
is  called  the  ratio  of  similitude  of  the  polygons. 
B 


'0' 


The  polygons  A  B  C  D  E  and  A'  I?  C  D'  E'  are 
similar  if  Z.  A  =  Z  ^',  Z  /?  =  Z  /?',  etc.,  and 
^^  £C  CZ) 

^'  ff  ~'~  B'  C  ~~  ~~  C  U  '  € 

The  ratio       „  ^.    is  the  ratio  of  similitude  of  the  poly- 


A'  B' 


gons. 


Ex.  133.  If  two  opposite  sides  of  a  quadrilateral  in- 
scribed in  a  circle  be  equal,  ihe  other  two  sides  are  par- 
allel. 

Ex.  134.  The  locus  of  the  middle  points 
of  all  chords  which  pass  through  a  given 
point,  is  a  circle  whose  diameter  is  the  line 
connecting  the  given  point  and  the  center 
of  the  circle. 

Prove  that  the  circle  described  on  O  S  is 
the  required  locus. 

Ex.  135.  Through  one  of  the  points  common  to  two 
intersecting  circumferences,  draw  the  diameters  of  the 
circles  and  prove  that  the  line  connecting  the  other  ex- 
tremities of  the  diameters  passes  through  the  other  point 
of  intersection  of  the  circumferences. 


154*  PLANE   GEOMETRY. 

PROPOSITION  XV.    THEOREM. 

230.  Two  triangles  which,  cure  mutually  equian- 
gular are  similar. 


A1 


Let  ABC  and  A  B'  C'  1>e  two  triangles  in  which  the 
angle  A  =  the  angle  A',  tJie  angle  B  =  toe  angle  B'  and 
the  angle  C  =  the  angle  C'  . 

To  prove  that  the  triangles  ABC  and  A'  B'  C'  are 
similar.  . 

SUG.  1.  Place  the  A  A'  B'  C  upon  the  A  A  B  C,  so 
that  A'  B'  lies  upon  A  B  with  A'  upon  A  and  B'  upon  M. 

SUG.  2.  What  direction  will  A'  C'  take  ?     Why  ? 

SUG.  3.  Where  will  the  point  C'  fall  ? 

SUG.  4.  What  relation  of  position  does  M  N  sustain  to 
BC1  Why? 

K    AM       AN     _ 
SUG.  5.  —r-fi  =  -A~T'     ^lve  autn- 

A  £>         A  C 

A 
SUG.  6.  In  the  same  way  compare  the  ratios 


Jj 

A        BC 

and    -p-g.. 

SUG.  7.  Apply  definition  of  similar  polygons  and  draw 
conclusion. 

Therefore  - 


PROPORTION  —  SIMILAR    POLYGONS. 


155 


231.  COROLLARY  I.     If  two  triangles  have  two  angles 
of  one  respectively  equal  to  two  angles  of  the  other,  the 
triangles  are  similar. 

232.  COROLLARY  II.    Two  right  triangles  which  have 
an   acute  angle  of  one  equal  to  an  acute  angle  of  the 
other,  are  similar. 


Ex.  136.  If  the  sum  of  the  opposite 
angles  of  a  quadrilateral  equals  two  right 
angles,  prove  that  a  circumference  is 
possible  through  the  four  vertices. 

SUG.  A  circumference  may  be  de- 
scribed through  three  vertices,  as  A,  B 
and  C.  If  it  does  hot  pass  through  £>,  it  must  cut  the 
line  C  D,  or  C  D  extended,  as  at  M.  Compare  the  sum 
of  Zs  B  and  C  M  A  with  two  rt.  Zs.  Also,  the  sum  of 
Zs  B  and  C  D  A  with  two  rt.  Zs. 


Ex.  137.  If  a  quadrilateral  be  circum- 
scribed about  a  circle,  the  sum  of  one 
pair  of  opposite  sides  is  equal  to  the  sum 
of  the  other  pair. 


Ex.  138.  If  two  Circles  are  tangent,  and  two  secants 
be  drawn  through  the  point  of  con- 
tact,  the  chords  joining  the  inter- 
sections of  the  secants  and  the  cir- 
cumferences are  parallel. 

Prove  A  D  parallel  to  B  C. 

SUG.  Draw  the  common  tangent, 
M  N.     Compare  Z.  A  O  M  with   Z  C  O  N.     Compare 
Z  A  OMwith  Z.  ADO\  also,  ZCONvtiih 


156  PIWVNE    GEOMETRY. 

PROPOSITION  XVI.     THEOREM. 

233.  If  two  triangles  have  an  angle  of  one  equal  to 
an  angle  of  the  other,  and  the  sides  including  the 
equal  angles  proportional,  the  triangles  are  similar. 


In  the  triangles  ABC  and  A'  B'  C'  let  the  angle  A 

A      T> 

equal  the  angle  A'  and  let  the  ratio         ^r    equal  the  ra- 

AC 

tio      .,  ,„ . 

A    C 

To  prove  that  the  triangles  ABC  and  A'  B'  C'  are 
similar. 

SuG.  1.  Place  the  A  A  B  C  upon  the  A  A'  B'  C',  with 
A  upon  A't  and  A  B  and  A  C  upon  A'  B'  and  A'  C'  re- 
spectively. Why  is  this  possible  ?  * 

SuG.  2.  Where  will  B  and  C  fall  ? 

SUG.  3.  What  relation  of  position  will  the  line  B  C 
sustain  to  B'  C  ?  Why  ? 

SUG.  4.  How  do  the  ^s  B  and  C  compare  with  the 
^s  B'  and  C'  respectively  ? 

SUG.  5.  See  Prop.  XV. 

Therefore • 


PROPORTION  —  SIMILAR   POLYGONS.  157 

PROPOSITION  XVII.     THKOREM. 

234.  Two  triangles  whose  sides  are  proportional 
are  similar. 

* 


r^     ~—          BO          ^O 
C  ~AJW  ~  W~W  ~  ATV' 

To  prove  that  the  triangles  ABC  and  A'  &  C'  are 

similar. 

SUG.  1  On  A  B,  lay  off  A  M  equal  to  A'  B'  and  on 
A  C,  lay  off  A  N  equal  to  A'  C'.  Connect  Maud  N. 

A  B  A'  /?' 

SUG.  2.  Compare  the  ratios  and     p,       .    Give 

yy  0  -£*    ^ 

auth. 

SUG.  3.  Compare   -^-^    with     „       ;      /.  *.,   compare 

-^-^    with      -,  „   .     Give  auth. 

SUG.  4.  Compare  B'  C  and  M  N. 

SUG.  5.  Then,  how  do  As  A'  B'  C'  and  A  M  N 
compare  ?  Why  ? 

SUG.  6.  The  line  M  N  divides  A  B  and  A  C  propor- 
tionally. Why  ?  See  hyp. 

SUG.  7.  How  is  the  A  A  MN  related  to  the  A  A  B  C? 

SUG.  8.  Then,  how  are  the  As  A'  B'  C'  and  A  B  C 
related  ? 

Therefore 


158 


PLANE   GEOMETRY. 


PROPOSITION  XVIII.     THEOREM. 

235.  The  ratio  of  homologous  altitudes  of  similar 
triangles  equals  the  ratio  of  similitude  of  the  tri- 
angles. 


B 


Let  A  Mand  E  N  be  homologous  altitudes  in  tJie  similar 
triangles  ABC  and  E  FG9  and  let  -^-~,  represent  the 
ratio  of  similitude  of  the  triangles. 

.  AM        AB 
To  prove  that    „  Ar   =  -=r-^. 
jc,  jtv  E,  r 

SuG.  1.  The  As  A  MB  and  E NF are  rt.  As.     Why  ? 

SuG.  2.  In  the  rt.  As  A  MB  and  E  N  F,  compare 
Z  B  with  Z  F. 

SuG.  3.  Then,  how  are  the  As  A  M B  and  E  NF  re- 
lated? See  Art.  232. 

SuG.  4.  Compare  the  ratio        „.   with  the  ratio 

Give  auth. 

Therefore 


Ex.  139.  Through  a  given  point,  to  draw  a  straight  line 
so  that  the  portion  of  it  intercepted  between  two  given  inter- 
secting lines  shall  be  divided  at  the  point  in  a  given  ratio. 

SUG.  Through  the  point  draw  a  line  parallel  to  one  of 
the  given  lines. 


PROPORTION  —  SIMILAR   POLYGONS. 


159 


PROPOSITION  XIX.     THEOREM. 

236.  //  two  polygons  are  composed,  of  the  same 
number  of  triangles,  similar  each  to  each  and  simi- 
larly placed,  the  polygons  are  similar. 


C' 


Let  the  polygons  A  B  C  D  E  F  and  A'  B'  C'  D'  E'  F1 
be  composed  of  the  same  number  of  triangles,  similar 
each  to  each  and  similarly  placed,  ABC  being  similar  to 
A'  B'  C',  and  so  on  for  the  other  triangles. 

To  prove  that  the  polygons  are  similar. 

SUG.  1.  Compare  the  Z  A  with  the  Z  A',  Z.  B  with 
Z.B' ,  etc. 

SUG.  2.  Compare  the  ratio       ,   „,    with 


SUG.  3.  Compare  the  ratio 


B  C 
B1  C 


with 


A'C 


also 


the  ratio 


with 


A  C 
A'C' 


B  C 

Now,  compare         ,  „, 


with 


CD 


CD" 

SUG.  4.  In  a  similar  manner  compare  the  ratios  of 
other  corresponding  sides  in 'the  two  polygons. 

SUG.  5.  Apply  the  definition  of  similar  polygons,  and 
draw  the  conclusion. 

Therefore  • 


160 


PLANE    GEOMETRY. 


PROPOSITION  XX.    THEOREM. 

237.  Converse  of  Prop.  XIX.  Two  similar  polygons 
may  be  divided  into  the  same  number  of  triangles, 
similar  each  to  each  and  similarly  placed. 


Let  the  polygons  AB  C  D  EF  and  A'  B'  €'  D'  E'  F' 
be  similar,  and  let  all  possible  diagonals  be  drawn  from 
corresponding  vertices  A  and  A'. 

To  prove  that  the  triangles  in  one  polygon  are  similar  to 
those  in  the  other  polygon. 

SUG.  1.  Compare  the  A  A  B  C  and  A'  B'  C'     (See 

Prop.  XVI. 

AC  B  C 

SUG.  2.  Compare  the  ratios 

CD         ,      B  C 
CD' 
CD 


and 


also, 


the  ratios 


A'C  B'C 

A  C 
Compare  the  ratios     ,  _, 

^i     C 


SUG.  3.  Compare  Z.  A  CD  with  Z.  A1  CD1. 
SUG.  4.  Compare  the  As  A  CD  and  A'  C  D' . 
SUG.  5.  In   a  similar  way  compare  the  other  corre- 
sponding As  in  the  two  polygons. 
Therefore 


PROPORTION — SIMILAR    POLYGONS.  161 

PROPOSITION  XXI.     THEOREM. 

238.  The  ratio  of  the  perimeters  of  two  similar 
polygons  equals  the  ratio  of  similitude  of  the  polygons. 

B> 


perimeter  ABODE 
To  prove  that  the  ratio    -  —r-      —  -—-    equals 

perimeter  A  B'  C  D  E      * 
A  ~fi 
the  ratio  of  similitude  . 

SUG.  See  proposition  X,  and  give  the  demonstration. 
Therefore 


Ex.  140.  Straight  lines 
drawn  through  an)7  point 
will  intercept  proportional 
segments  upon  two  parallel 
lines. 

Prove  that  -    _ 


-  /B 

~  FN  >e 

AC  C  P 

SUG.  Compare  each  of  the  ratios  -ir-=  and  -=—  =,  with 

Jo  D  D  f1 

the  ratio 


Complete  the  demonstration. 

11—  Geo. 


162  PLANE    GEOMETRY. 

PROPOSITION  XXII.    THEOREM. 

239.  If  a  perpendicular  be  dropped  to  the  hypote- 
nuse from  the  vertex  of  a  right  angle  in  a  right  tri- 
angle: 

!•  The  triangles  thus  formed  are  similar  to  each 
other  and  to  the  whole  triangle. 

II.  The  perpendicular  is  a  mean  proportional  be- 
tween the  segments  of  the  hypotenuse. 

III.  Each  side  is  a  mean  proportional  between  the 
hypotenuse  and  the  segment  adjacent  to  that  side. 


Let  ABC  represent  a  right  triangle,  right  angled  at~A9 
and  let  AM  represent  a  perpendicular  drawn  from  A  to 
tlie  hypotenuse  B  C. 

I.  To  prove  that  the  triangles  A  M  C,  A  MB  and 
B  A  C  are  similar. 

SUG.  1.  In  the  rt.  As  A  M  C  and  A  MB,  compare 
Z  CA  MvfiihZB. 

SUG.  2.  See  Art.  232. 

,    SUG.  3.  In  the  rt.  As  A  M  C  and  B  A  C,  notice  that 
Z  C  is  common.     See  Art.  232. 

TT       T  ,/    ,     CM         AM 

II.  To  prove  that    -^  =  ^-^. 

SUG.  Deduce  this  proportion  from  the  similar  AsAMC 
zn&AMB. 


PROPORTION  —  SIMILAR   POLYGONS.  163 


.,    .  CB      AB  CB      CA 

III.      To  prove  that—  B= 


SUG.  Deduce  the  first  proportion  from  the  similar 
As  ABC  and  ABM.  Deduce  the  second  proportion 
from  the  similar  As  A  B  Cand  ACM. 

Therefore  - 

*     PROPOSITION  XXIII.    THEOREM. 

240.  If  two  chords  of  a  circle  intersect,  the  ratio  of 
either  segment  of  the  first  to  either  segment  of  the  sec- 
ond equals  the  ratio  of  the  remaining  segment  of  the 
second  to  the  remaining  segment  of  the  first. 


B 


Let  the  chords  A  B  and  C  D  intersect  at  X. 

AX       CX 


To  prove  that 


D  X    ~  B  X' 


SUG.  1.  Connect  A  and  C\  also,  B  and  D. 

SUG.  2.  Compare  Z.  A  with  Z  D\  also,  Z.  C  with 
Z  B.  (Art.  177.) 

SUG.  3.  Compare  As  A  X  C  and  D  XB  with  respect 
to  form.  (Art.  230.) 

SUG.  4.  Complete  the  demonstration. 

Therefore 


164  PLANE    GEOMETRY. 

PROPOSITION  XXIV.    THEOREM. 

241.  If  two  secants  intersect  without  a  circle,  the 
ratio  of  the  first  secant  to  the  second  equals  the  ratio 
of  the  external  segment  of  the  second  to  the  external 
segment  of  the  first. 


Let  A  B  and  A  C  represent  two  secants  meeting  at  A. 

To  prove  that    ±»c=lfD. 

SuG.  1.  Connect  D  and  C\  also,  B  and  E. 
SUG.  2.  Compare  the  As  A  D  C  and  A  E  B. 
SUG.  3.  Complete  the  demonstration. 
Therefore 


^      n/M     T,   A        C  ...   A  +  B       C+D 

Ex.  141.  If   ^  =  -^ ,  prove  that  -=j—^  =  ^—D. 

Are  A,  B,  C  and  D  restricted  to  being  numbers,  or  may 
they  be  quantities  ?  If  quantities,  are  they  all  of  the 
same  kind,  or  may  they  differ  in  kind  ? 

Ex.  142.  Construct  a  triangle  having  a  given  base,  a 
given  altitude  and  a  given  vertical  angle. 

SUG.  Solve  by  intersection  of  two  loci. 


PROPORTION  —  SIMILAR    POLYGONS.  165 

PROPOSITION  XXV.     THEOREM. 

242.  //'  a  secant  and  a  tangent  meet  without  a 
circle,  the  tangent  is  a  mean  proportional  between  the 
secant  and  its  external  segment. 


Let  A  C  represent  a  tangent,  and  O  C  a  secant  meeting 
the  tangent  at  tfie  point  C. 

O  C      AC 
To  prove  that    ^-j^. 

SUG.  1.  Connect  A  and  O\  also,  A  and  B. 
SUG.  2.   Compare  the  As  A  B  C  and  O  A  C  with  re- 
spect to  form. 

SUG.  3.  Complete  the  demonstration. 
Therefore 


Ex.  143.  What  is  the  locus  of  the  vertex  of  the  right 
angle  of  a  right-angled  triangle  constructed  upon  a  given 
line  as  a  hypotenuse?  (See  Ex.  122.) 

Ex.  144.  Construct  a  triangle  whose  base  and  vertical 
angle  are  given,  and  whose  vertex  is  at  a  given  distance 
from  the  middle  point  of  the  base. 

SUG.  Solve  by  intersection  of  two  loci. 


166  PLANE    GEOMETRY. 

PROPOSITION  XXVI.     PROBLEM. 

243.  To  divide  a  given  straight  line  into  any  num- 
ber of  equal  parts. 


K 


Let  A  B  be  the  given  straight  line. 

To  divide  A  B  into  any  number  of  equal  parts — for  ex- 
ample,  four. 

SUG.  1.  Through  A,  draw  an  indefinite  straight  line 
making  any  convenient  Z.  with  A  B. 

SUG.  2.  Take  any  convenient  unit  of  length  and  lay  it 
off  four  times  on  the  indefinite  line. 

SUG.  3.  Connect  the  last  point  of  division,  F,  with  the 
point  B,  and  draw  ||  s  to  FB  through  the  other  points  of 
division,  C,  D  and  E. 

SUG.  4.  Why  will  these  ||  s  divide  A  B  into  equal 
parts  ? 


Ex.  145.  With  a  given  fixed  base,  and  a  given  verti- 
cal angle  of  a  triangle,  find  the  locus  of  the  vertex.  See 
exercise  121. 

Ex.  146.  Construct  upon  a  given  line,  as  a  chord,  a 
segment  of  a  circle  which  shall  contain  a  given  angle. 

SuG.  Construct  at  the  ends  of  the  given  line,  and  upon 
it,  angles  whose  sum  is  the  supplement  of  the  given  angle. 


PROPORTION — SIMILAR   POLYGONS.  167 

PROPOSITION  XXVII.     THEOREM. 

244.  To  divide  a  given  straight  line  into  parts  pro- 
portional to  any  given  lines. 


F 


B 


Let  A  B  be  the  given  straight  line. 

To  divide  A  B  into  parts  such  that   —^r  =  — ry- ,  etc. 

M          N 

SUG.  1.  Through  A,  draw  an  indefinite  straight  line 
making  any  convenient  ^  with  A  B. 

SUG.  2.  On  this  indefinite  line  lay  off,  in  succession, 
the  lines  M,  N,  R,  etc. 

SUG.  3  Connect  the  last  point  of  division  with  the 
point  £,  and  draw  ||  s  from  the  other  points  of  division, 
C,  Dt  etc. 

SUG.  4.  Why  will  these  ||  s  divide  A  B  into  parts  pro- 
portional to  M,  N,  R,  etc.? 

245.  A  fourth  proportional  to  three  given  quanti- 
ties is  the  fourth  term  of  a  proportion  whose  first  three 
terms  are  the  three  given  quantities  taken  in  order.  If 
the  three  given  quantities  are  6  ft.,  12  ft.  and  18  ft.,  the 
fourth  proportional  is  36  ft. ;  if  the  given  quantities  are 
12  ft.,  6  ft.  and  18  ft.,  the  fourth  proportional  is  9  ft.;  if 
the  given  quantities  are  18  ft.,  6  ft.  and  12  ft.,  the  fourth 
proportional  is  4  ft. 


168  PLANE    GEOMETRY. 

PROPOSITION  XXVIII.     PROBLEM. 

246.  To  find  a  fourth  proportional  to  three  given 
lines. 


•B 


C D 

B P 


Let  A  B,  C  D  and  E  F  be  three  given  lines. 

To  find  a  fourth  proportional  to  these  three  lines. 

SUG.  1.  In  the  figure  in  Art.  226,  notice  that  E  C  is  a 
fourth  proportional  to  A  D,  D  B  and  A  E. 

SUG.  2.  Construct  a  figure  in  which  one  of  the  lines 
shall  be  a  fourth  proportional  to  A  B,  CD  and  E  F. 

SUG.  3.  Work  out  at  least  three  solutions  of  this  prob- 
lem. 

247.  A  third  proportional  to  two  given  quantities  is 
the  fourth  term  of  a  proportion  whose  first  term  is  one  of 
the  given  quantities,  and  whose  second  and  third  terms 

are  each  the  second  of  the  given  quantities.     If  7-  =  — , 
c  is  a  third  proportional  to  a  and  b. 


Ex.  147.  Give  a  solution  of  Prop.  XXXII,  Chap.  II, 
by  using  the  truth  that  if  two  lines  are  perpendicular  to 
the  same  line  they  are  parallel. 


PROPORTION  —  SIMILAR    POLYGONS.  169 

PROPOSITION  XXIX.     PROBLEM. 
248.   To  find  a  third  proportional  to  two  given  lines. 


A. 


C D 

Let  A  J5  and  C  D  be  two  given  lines. 

To  find  a  third  proportional  to  A  B  and  C  D. 

SUG.  1.  See  suggestions  to  the  preceding  proposition, 
and  make  the  construction. 

SUG.  2.  Make  out  at  least  two  solutions. 

PROPOSITION  XXX.     PRJDBLEM. 

249.   To  find  a  mean  proportional  between  two  given 
lines. 


A. 


C D 

Let  A  B  and  C  D  be  two  given  lines. 

To  find  a  mean  proportional  between  A  B  and  C  D. 

SUG.  1.  See  suggestions  to  Prop.  XXII,  II  and  make 
the  construction. 

SUG.  2.  Make  out  two  or  more  solutions.  Every  prop- 
osition which  involves  a  mean  proportional  should  sug- 
gest a  solution  to  the  problem. 


Ex.  148.  Prove    Prop.   XVIII,  Chap.  Ill,   by  taking 

f~* 

-~  as  the  ratio  of  similitude  of  the  triangles. 


170  PLANE   GEOMETRY. 

PROPOSITION  XXXI.     PROBLEM. 

250.  To  construct,  upon  a  given  line  as  a  ~base,  a 
triangle  similar  to  a  given  triangle. 


B 


C 
Let  A  B  C  be  a  given  triangle,  and  E  Ga  given  line. 

To  construct,  upon  E  G  as  a  base,  a  triangle  similar  to 
the  triangle  ABC. 

SUG.  1.  If  B  C  be  regarded  as  the  base  of  the  A  A  B  C, 
how  must  the  /_  at  E,  in  the  required  A  ,  compare  with 
the  Z  B  ? 

SUG.  2.  How  must  the  ^  at  G  compare  with  the  Z.  C? 

SUG.  3.  Make  the  required  construction. 


Ex.  149.  If  two  parallel  lines 
are  cut  proportionally  by  a  set 
of  secant  lines,  prove  that  the 
secant  lines  all  pass  through  a 
common  point. 

_-  AC       CE 


that  the  lines  B  A,  D  C,  F  E, 
etc.,  intersect  at  a  common 
point. 

SUG.  Extend  two  of  them,  B  A  and  D  C,  until  they 
meet  at  O.  Connect  O  and  E,  and  extend  O  E  to  the 
other  of  the  parallels.  Determine  the  relative  position 


of  the  lines   O  M   and  E  F.      ~~  =  -      Why? 

J3  D 

AC        CE 

why? 


PROPORTION  — SIMILAR   POLYGONS.  171 

PROPOSITION  XXXII.     PROBLEM. 

251.  To  construct  a  polygon  similar  to  a  given  poly- 
gon and  liaving  a  given  line  as  one  side. 


Let  MN P  8 K  be  a  given  polygon,  and  M.'  XT  a  given 
straight  line. 

To  construct \  with  M'  N'  as  one  side,  a  polygon  similar 
toMNPSR. 

SUG.  1.  Let  M'  N'  be  homologous  to  the  side  M  N, 
and  from  M  draw  all  possible  diagonals  of  the  given 
polygon. 

SUG.  2.  On  M'  N'  construct  a  A,  M'  N1  P* ,  similar 
to  the  A  MNP.  Give  auth. 

SUG.  3.  On  M'  P'  construct  a  A  similar  to  the 
A  M  P  S.  Continue  the  construction. 


PROPOSITIONS   IN   CHAPTER  III. 


PROPOSITION  I. 

In  a  proportion,  all  of  whose  terms  are  numbers,  the  product  of 
the  means  equals  the  product  of  the  extremes. 

PROPOSITION  II. 

If  the  product  of  two  numbers  equals  the  product  of  two  other 
numbers,  the  factors  of  one  product  may  be  made  the  means,  and  the 
factors  of  the  other  product  the  extremes,  of  a  proportion. 

PROPOSITION  III. 
To  determine  a  mean  proportional  between  two  given  numbers. 

PROPOSITION  IV. 

If,  in  several  successive  ratios,  the  consequent  of  the  first  equals 
the  antecedent  of  the  second,  the  consequent  of  the  second  equals  the 
antecedent  of  the  third,  etc. ,  the  ratio  of  the  antecedent  of  the  first  to 
the  consequent  of  the  last  equals  the  product  of  the  ratios. 

PROPOSITION  V. 

Both  terms  of  a  ratio  may  be  multiplied  by  any  number  without 
changing  the  value  of  the  ratio. 

PROPOSITION  VI. 

If  four  numbers,  or  like  quantities,  are  in  proportion,  the  ratio  of 
the  first  to  the  third  equals  the  ratio  of  the  second  to  the  fourth. 

PROPOSITION  VII. 

If  four  numbers,  or  quantities,  are  in  proportion,  the  ratio  of  the 
second  to  the  first  equals  the  ratio  of  the  fourth  to  the  third. 

PROPOSITION  VIII. 

If  four  numbers,  or  quantities,  are  in  proportion,  the  ratio  of  the 
first  plus  the  second,  to  the  second,  equals  the  ratio  of  the  third  plus 
the  fourth,  to  the  fourth. 


PROPORTION  —  SIMILAR    POLYGONS.  173 


PROPOSITION  IX. 

If  four  numbers,  or  quantities,  are  in  proportion,  the  ratio  of  the 
first  minus  the  second,  to  the  second,  equals  the  ratio  of  the  third 
minus  the  fourth,  to  the  fourth. 

PROPOSITION  X. 

In  a  series  of  equal  ratios,  all  of  whose  terms  are  numbers  or 
quantities  of  the  same  kind,  the  ratio  of  the  sum  of  the  antecedents  to 
the  sum  of  the  consequents  equals  any  one  of  the  ratios. 

PROPOSITION  XI. 

If  three  terms  of  one  proportion  are  respectively  equal  to  the  three 
corresponding  terms  of  another  proportion,  the  fourth  terms  are 
equal. 

PROPOSITION  XII. 

If  parallel  lines  intercept  equal  segments  on  one  transversal,  they 
intercept  equal  segments  on  all  transversals. 

PROPOSITION  XIII. 

If  a  line  is  parallel  to  the  base  of  a  triangle,  the  ratio  of  the  seg- 
ments of  one  side  equals  the  ratio  of  the  segments  of  the  other  side. 

PROPOSITION  XIV. 

If  a  line  divides  two  sides  of  a  triangle  proportionally,  it  is  parallel 
to  the  base. 

PROPOSITION  XV. 
Two  triangles  which  are  mutually  equiangular  are  similar. 

PROPOSITION  XVI. 

If  two  triangles  have  an  angle  of  one  equal  to  an  angle  of  the  other, 
and  the  sides  including  the  equal  angles  proportional,  the  triangles 
are  similar. 

PROPOSITION  XVII. 
Two  triangles  whose  sides  are  proportional  are  similar. 


174  PLANE    GEOMETRY. 


PROPOSITION  XVIII. 

The  ratio  of  homologous  altitudes  of  similar  triangles  equals  the 
ratio  of  similitude  of  the  triangles. 

PROPOSITION  XIX. 

If  two  polygons  are  composed  of  the  same  number  of  triangles, 
similar  each  to  each  and  similarly  placed,  the  polygons  are  similar. 

PROPOSITION  XX. 

CONVERSE  OF  PROP.  XIX.  Two  similar  polygons  may  be  divided 
into  the  same  number  of  triangles,  similar  each  to  each,  and  similarly 
placed. 

PROPOSITION  XXI. 

The  ratio  of  the  perimeters  of  two  similar  polygons  equals  the  ra- 
tio of  similitude  of  the  polygons. 

PROPOSITION  XXII. 

If  a  perpendicular  be  dropped  to  the  hypotenuse  from  the  vertex  of 
a  right  angle  in  a  right  triangle: 

I.  The  triangles  thus  formed  are  similar  to  each  other  and  to  the 
whole  triangle. 

II.  The  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

III.  Each  side  is  a  mean  proportional  between  the  hypotenuse 
and  the  segment  adjacent  to  that  side. 

PROPOSITION  XXIII. 

If  two  chords  of  a  circle  intersect,  the  ratio  of  either  segment  of 
the  first  to  either  segment  of  the  second  equals  the  ratio  of  the  re- 
maining segment  of  the  second  to  the  remaining  segment  of  the  first. 

PROPOSITION  XXIV. 

If  two  secants  intersect  without  a  circle,  the  ratio  of  the  first  secant 
to  the  second  equals  the  ratio  of  the  external  segment  of  the  second  to 
the  external  segment  of  the  first. 

PROPOSITION  XXV. 

If  a  secant  and  a  tangent  meet  without  a  circle,  the  tangent  is  a 
mean  proportional  between  the  secant  and  its  external  segment. 


PROPORTION  —  SIMILAR   POLYGONS.  175 

PROPOSITION  XXVI. 
To  divide  a  given  straight  line  into  any  number  of  equal  parts. 

PROPOSITION  XXVII. 

To  divide  a  given  straight  line  into  parts  proportional  to  any  given 
lines. 

PROPOSITION  XXVIII. 
To  find  a  fourth  proportional  to  three  given  lines. 

PROPOSITION  XXIX. 
To  find  a  third  proportional  to  two  given  lines. 

PROPOSITION  XXX. 
To  find  a  mean  proportional  between  two  given  lines. 

PROPOSITION  XXXI. 

To  construct,  upon  a  given  line  as  a  base,  a  triangle  similar  to  a 
given  triangle. 

PROPOSITION  XXXII, 

To  construct  a  polygon  similar  to  a  given  polygon  and  having  a 
given  line  as  one  side. 


CHAPTER  IV. 
COMPARISON  AND  MEASUREMENT  OF  POLYGONS. 


DEFINITION. 

252.  The  area  of  a  surface  is  its  ratio  to  some  selec- 
ted unit  of  measure,  coupled  with  the  unit  of  measure. 
(See  Arts.  159  and  160.) 

If  m  denotes  the  unit  of  measure  for  determining  the 

A 

area  of  a  surface  A,  and  if  the  ratio  --  equals  b  (article 

161),  the  area  of  the  surface  A  is  bm  ;  i.  e.,  b  times  the 
unit  m. 

The  unit  ef  measure  used  to  express  the  area  of  sur- 
faces is  a  square,  whose  side  is  a  given  linear  unit;  as  a 
square  inch,  a  square  foot,  etc. 

PROPOSITION  I.     THEOREM. 

253.  Two  parallelograms  having  equal  bases  and 
equal  altitudes  are  equal  in  area* 

B      A'  -m*  B'A' 


Let  A  B  CD  and  A'  B'  €'  D'  be  two  parallelograms 
having  D  C  and  D'  C'  equal,  and  the  altitudes  m  n  and 
m'  n'  equal. 

To  prove  that  the  area  of  A  C  equals  the  area  of  A'  Cf. 


MEASUREMENT   OF    POLYGONS. 


177 


SUG.  Place  the  EJ  A  C  upon  the  £7  A'  C',  D  C  upon 
U  C1 ,  and  compare  the  parts  external  to  each  other. 
Therefore 

PROPOSITION  II.    THEOREM. 

254.  If  two  rectangles  have  equal  altitudes,  the  ra- 
tio of  their  areas  equals  the  ratio  of  their  bases. 

A  RE  F 


Let  A  C  and  E  G  be  two  rectangles  having  equal  alti- 
tudes, D  A  and  HE. 

A   area  of  A  C        DC 
To  prove  that  -    — J '    _  _  =-  ^—^ . 
area  of  EG        HG 

CASE  I.      When  D  C  and  H  G  are  commensurable. 

SUG.  1.  Measure  D  Cand  H  G  by  some  common  unit 
of  measure.  Let  the  unit  be  contained  m  times  in  D  C, 
and  n  times  in  H  G. 

SUG.  2.  What  is  the  ratio  ofDCto&G* 

SUG.  3.  Through  the  points  of  division  erect  J_s  to 
D  C  and  H  G,  and  extend  the  JLs  to  the  secondary  bases, 
thus  dividing  A  C  and  E  G  into  rectangles. 

SUG.  4.  How  many  rectangles  in  A  C?  How  many  in 
H  G  ?  How  do  these  smaller  rectangles  compare  in  size  ? 
Why? 

SUG.  5.  What  does  the  ratio  of  the  areas  of  the  rect- 
angles A  C  and  E  G  equal  ? 

SUG.  6.  Compare  the  ratio  of  the  areas  with  the  ratio 
of  the  bases. 
12— Geo. 


178 


PLANE   GEOMETRY. 


CASK  II.      When  the  bases  are  incommensurable. 
ARM  BE  G 


SUG.  1.  Take  any  measure  of  K  /and  lay  it  off  on  D  C 
as  many  times  as  possible.  There  must  be  a  remainder 
less  than  one  of  these  parts.  Why  ?  Suppose  the  meas- 
ure of  K  /is  contained  in  D  C  a  certain  number  of  times 
with  a  remainder  D  N.  Erect  N  M  J_  to  D  C,  at  the 
point  N. 

SUG.  2.  Compare  the  ratio  of  the  areas  ofMC  and  E  I 
with  the  ratio  of  N  C  and  K  I. 

SUG.  3.  Use,  now,  a  unit  of  measure  smaller  than  D  N, 
and  let  the  remainder  now  be  D  S.  Erect  S  R  _L  to  D  C 
at  S,  and  compare  the  ratio  of  the  areas  of  R  C  and  E  I 
with  the  ratio  of  5  C  and  K  I. 

SUG.  4.  By  continually  decreasing  the  unit  of  measure 
the  two  ratios  compared  are  variables.  The  limit  of  the 

area  of      ° 


first  ratio  is  T  . 

area  of  E  I 

ond  ratio  is  -=^rj  .     Why  ? 


Why  ?    The  limit  of  the  sec- 


SUG.  5.  What  relation  exists  between  the  variables  as 
they  approach  their  limits  ?  (See  Siig.  3  ) 

SUG.  6.  What  relation  exists  between  the  limits  ? 
Therefore  - 

255.  COROLLARY.  If  two  rectangles  have  equal  bases 
the  ratio  of  their  areas  equals  the  ratio  of  their  altitudes. 

Note  i.  —  Compare  the  demonstration  of  Prop.  II  with  that  of 
Prop.  XIX,  Chap.  II.  What  other  proposition  is  demonstrated  in  a 
similar  manner  ? 


MEASUREMENT  OF  POLYGONS. 


179 


Note  2. —  Hereafter  a  ratio  like    -^—r     will  be  written  in  the 

area  &  I 

A  C 
form   -F-r. 


PROPOSITION  III    THEOREM. 

256.  The  number  of  units  of  area  in  any  rectangle 
is  equal  to  the  product  of  the  number  of  linear  units  in 
the  base  and  altitude. 


"Let  A  represent  a  rectangle,  U  a  unit  of  measure  for 
area,  and  u  the  linear  unit,  viz.,  a  side  of  the  square  U. 
Let  u  be  contained  a  times  in  m  and  b  times  in  n. 

To  prove  that  A  contains  the  unit  Uy  a  X  b  times. 

SUG.  1.  Construct  a  rectangle  B  whose  altitude  is  n 
and  whose  base  is  u. 

SUG.  2.  What  is  the  ratio  of  A  to  B1  (Prop.  II.)  Re- 
duce the  answer  to  the  simplest  form. 

SUG.  3.  What  is  the  ratio  of  B  to  £/?  Reduce  as  be- 
fore. 

SUG.  4.  What,  then,  is  the  ratio  of  A  to  Ut  (Art.  206.) 

Therefore 


Ex.  150.  In  a  given  line  determine  a  point  which  is 
equally  distant  from  two  given  points  not  in  the  line. 

Note. —  In  this,  and  as  many  other  problems  of  construction  as  pos- 
sible, employ  the  intersection  of  loci. 


180 


PLANE    GEOMETRY. 


MODEL. 

PROPOSITION  III.     THEOREM. 

257.  The  number  of  units  of  area  in  any  rectangle 
is  equal  to  the  product  of  the  number  of  linear  units  in 
the  base  and  altitude. 


A 

n 

u. 

B 

m. 

n. 

Let  A  represent  a  rectangle,  V  a  unit  of  measure  for 
area,  and  u  the  linear  unit,  viz.,  a  side  of  the  square  V. 
Let  u  be  contained  a  times  in  m  and  b  times  in  n. 

To  prove  that  A  contains  the  unit  U,  a  x  b  times. 

Construct  the  rectangle  B,  whose  altitude  is  n  and 
whose  base  is  u. 

Since  A  and  B  have  equal  altitudes  —-=--.  (Prop- 
osition II.) 

But  since  u  is  contained  a  times  in  m,    •  -  =  a. 

u 
£ 

Hence,  •=-  =  a. 
n 

Ft          n, 
Since  B  and  U  have  equal  bases,    jj  —  — . 

But  since  u  is  contained  b  times  in  n,   —  =  b. 


n 

Hence,   -~ 


b. 


Since  -^  —  a,  and  -=-T  =  b 

£>  U 


MEASUREMENT  OF  POLYGONS.        181 

Therefore,    ^  =  a  x  b  (Art.  206). 

That  is,  A  contains  £/,  a  X  b  times. 

Therefore,  the  number  of  units  of  area  in  any  rectangle 
is  equal  to  the  product  of  the  number  of  linear  units  in 
the  base  and  altitude. 

258.  SCHOLIUM  I.     In  the  applications  of  this  theorem 
the  base  and  altitude  must  be  expressed  in  terms  of  the 
same  unit,  and  the  unit  of  area  must  be  the  square  whose 
side  is  the  linear  unit. 

259.  SCHOLIUM  II.     By  comparison  of  the  theorem 
with  the  definition  of  area  (Art.  252),  it  will  be  observed 
that  the  area  of  a  rectangle  is  the  product  of  the  measures 
of  the  base  and  altitude  joined  to  the  name  of  the  unit  of 
area. 

ILLUSTRATIONS.  If  the  base  of  a  rectangle  is  10  ft.  and 
the  altitude  6  ft.,  the  area  of  the  rectangle  is  (6  X  10) 
sq.  ft.  If  the  base  of  a  rectangle  is  10  ft.  and  the  altitude 
6  in.  (or  \  ft.),  the  area  of  the  rectangle  is  (£  X  10)  sq.  ft. 
(Sch.  I.) 

260.  SCHOLIUM  III.     The  expression  "  the  product  of 
the  base  and  altitude,"  is  a  common  abbreviation  for 
"the  product  of  the  measures  of  the  base  and  altitude 
joined  to  the  name  of  the  unit  of  area." 

' '  The  product  of  two  lines,  * '  or  ' '  the  square  of  a  line, ' ' 
must  not  be  interpreted  in  any  other  sense  than  that  just 
stated.  With  this  interpretation,  Prop.  Ill  is  usually 
stated  as  follows:  The  area  of  a  rectangle  is  equal  to  the 
product  of  the  base  and  altitude. 

The  numbers  which  represent  the  measures  of  the  lines 
may  be  integral,  fractional,  or  incommensurable. 


182 


PLANE   GEOMETRY. 


Note. —  It  may  make  the  thought  in  the  foregoing  demonstration 
clearer  to.illustrate  it  by  the  arithmetical  form  of  analysis. 

U  is  contained  in  B  the  same  number  of  times  that  u  is  in  n,  i.  *., 
b  times;  also,  B  is  contained  in  A  the  same  number  of  times  that  u  is 
in  m,  i.  e.t  a  times.  Now,  as  U  is  contained  b  times  in  Bt  and  B  is 
contained  a  times  in  A,  it  follows  that  U  is  contained  a  X  b  times  in  A. 

To  take  a  definite  case,  let  the  base 
and  altitude  be  measured  by  some 
linear  unit,  as  1  inch;  and  suppose 
this  unit  is  contained  5  times  in  the 
altitude  and  8  times  in  the  base, 
then,  as  seen  in  the  figure  at  the 
right,  there  are  8  columns,  with  5 
squares  in  each  column,  and  hence, 
in  all,  8  X  5  squares;  i.  e.t  40  square  inches. 

It  is  readily  seen,  by  constructing  a  figure,  that  if  the  unit  is  con- 
tained a  fractional  number  of  times  in  the  base  and  altitude  the  same 
rule  is  reached. 

QUERY.  Can  the  conclusion  of  the  proposition,  in  all  its  generality, 
be  drawn  from  the  method  in  common  use  in  arithmetic? 

PROPOSITION  IV.    THEOREM. 

26 1 .  The  area  of  a  parallelogram  equals  the  product 
of  its  base  and  altitude. 

SUG.  See  Props.  I  and  III,  and  demonstrate. 


Ex.  151.  Determine  a  point  equally  distant  from  two 
given  points,  and  at  a  given  distance  from  a  given  line. 

How  many  points  answer  the  conditions  of  the  prob- 
lem? 

Ex.  152.  Determine  a  point  equally  distant  from  two 
given  parallel  lines,  and  equally  distant  from  two  given 
points. 


MEASUREMENT  OF  POLYGONS.        183 

Ex.  153.  Determine  a  point  at  given  distances  from 
each  of  two  given  intersecting  lines. 

How  many  points  answer  the  conditions  of  the  prob- 
lem ? 

Ex.  154.  Determine  a  point  at  given  distances  from 
two  given  points. 


PROPOSITION  V.    THEOREM. 

262.  The  area  of  a  triangle  equals  one  half  the 
product  of  its  base  and  altitude. 


Let  ABC  represent  a  triangle,  AB  its  base  and  M  C 
its  altitude. 

To  prove  that  the  area  of  the  triangle  ABC  equals  one 
half  of  A  B  times  M  C. 

SUG.  1.  From  C,  draw  CD  equal  and  ||  to  A  B,  and 
connect  B  with  D.  What  kind  of  a  figure  is  A  D  ? 
Why? 

SUG.  2.  Compare  the  A  A  B  C  with  the  A  CD  B. 

SUG.  3.  Compare  the  A  A  B  C  with  the  EH  A  Z>,  in 
respect  to  magnitude;  also,  compare  bases  and  altitudes. 

SUG.  4.  What,  then,  is  the  area  of  the  A  A  B  C  in 
terms  of  A  B  and 

Therefore 


184 


PLANE   GEOMETRY. 


263,  COROLLARY.     If  two  triangles  have  equal  alti- 
tudes their  areas  have  the  same  ratio  as  their  bases. 


Then 


Give 


In  the  triangles  A  C  D  and  A1  C  Df,  let  the  altitudes 
A  B  and  A1  B'  be  equal. 

A  CD         \A  B  X   CD          CD 
A'  CD'     ''  \A'  B'  x  C' D'  ''''   CD'' 
auth. 

In  a  similar  manner,  prove  that  if  two  triangles  have 
equal  bases  their  areas  have  the  same  ratio  as  their  alti- 
tudes. 

264,  SCHOUUM.     For  the  interpretation  of  the  ex- 
pression, "  the  product  of  the  base  and  altitude,"  in  prop- 
ositions IV  and  V,  see  Art.  260. 

PROPOSITION  VI.    THEOREM. 

265.  The  area  of  a  trapezoid  equals  one  half  the 
product  of  its  altitude  by  the  swm  of  its  'bases. 


Let  A  B  CD  represent  a  trapezoid,  and  A  Mits  altitude. 


MEASUREMENT  OF  POLYGONS.        185 

To  prove  that  the  area  of  the  trapezoid,  A  B  C  D,  equals 
one  half  of  A  M  times  the  sum  of  A  B  and  D  C. 

SUG.  1.  Draw  the  diagonal,  B  D. 

SUG.  2.  In  the  A  D  C  B,  if  D  C  is  taken  as  the  base, 
what  is  the  altitude?  Then,  what  is  the  area  of  the 
A  D  C  B  equal  to  ? 

SUG.  3.  In  the  A  D  A  B,  if  A  B  is  taken  as  the  base, 
what  is  the  altitude?  Then,  what  is  the  area  of  the 
A  D  A  B  equal  to  ? 

SUG.  4.  What,  then,  is  the  area  of  the  trapezoid  A  BCD 
equal  to  ? 

Therefore 

Ex.  155.  The  ratio  of  the  squares  of  the  legs  of  a  right 
triangle  is  equal  to  the  ratio  of  the  segments  formed  by 
dropping  a  perpendicular  from  the  vertex  of  the  right 
angle  upon  the  hypotenuse. 

SUG.  1.  See  figure  in  Art.  239.  By  use  of  part  III  of 
the  proposition  in  Art.  239,  find  an  expression  for  the 
square  of  each  leg  of  the  triangle. 

SUG.  2.  What  is  the  ratio  of  the  squares  found  ? 

Ex.  156.  The  sum  of  the  squares  of  the  legs  of  a  right 
triangle  equals  the  square  of  the  hypotenuse. 

SUG.  1.  See  Sug.  1,  of  Ex.  155. 

SUG.  2.  Add  the  squares. 

Ex.  157.  Determine  a  point  at  a  given  distance  from  a 
given  point,  and  equally  distant  from  two  parallel  lines. 

How  many  points  answer  the  conditions  of  the  problem  ? 

Is  this  problem  always  possible  ? 

Ex.  158.  Determine  a  point  at  a  given  distance  from  a 
given  point,  and  equally  distant  from  two  given  intersect- 
ing lines. 

Show  when  there  are  four  points,  when  three,  when 
two,  when  one,  and  when  not  any. 


186 


PLANE   GEOMETRY. 


AREA  OF  A  POLYGON. 

266.   Various  methods  have  been  used  to  find  the  area  of  poly- 
gons.    Among  the  methods  used  the  following  may  be  noticed: 

From  any  vertex  of  the  polygon  draw  all 
possible  diagonals,  as  in  fig.  1,  at  the  right. 
The  polygon  is,  by  this  means,  divided  into 
triangles,  and,  if  the  bases  and  altitudes  of 
these  triangles  can  be  measured,  their  areas 
can  be  computed,  and  then,  by  addition, 
the  area  of  the  polygon  can  be  found. 

Another  method  is  to  draw  the  longest 
diagonal  of  the  polygon,  and,  from  the  ver- 
tices, drop  perpendiculars  upon  this  diag- 
onal, as  in  fig.  2,  at  the  right.  The  poly- 
gon is,  in  this  way,  divided  into  triangles 
and  trapezoids,  and,  if  the  bases  and  alti- 
tudes of  these  triangles  and  trapezoids  can 
be  measured,  their  areas  can  be  computed, 
and  then,  by  addition,  the  area  of  the  poly- 
gon can  be  obtained. 

Still  another  method  is  to  draw,  through 
any  vertex  of  the  polygon,  a  straight  line, 
exterior  to  the  polygon,  and,  from  the  ver- 
tices, drop  perpendiculars  upon  this  line,  as  in  fig.  3. 
In    this  way,  triangles   aud  trapezoids  are  formed, 
and,  if  the  various  bases  and  altitudes  can  be  meas- 
ured, the  areas  of  the  triangles  and  trapezoids  can  be 
computed,  and,  if  the  areas  of  the  parts  exterior  to 
the  polygon  be  subtracted  from  the  sum  of  the  other 
areas,  the  difference  gives  the  area  of  the  polygon.  FIQ  3 

The  method  just  described  is  the  one  by  which  surveyors  sometimes 
compute  irregular  areas  bounded  by  straight  lines. 

In  considering  the  area  of  a  field,  or  any  large  irregular  polygon, 
it  is  often  convenient  to  make  a  map  or  outline  of  the  polygon  accu- 
rately drawn  to  a  scale  as  large  as  convenient.  Then,  any  line  con- 
nected with  the  polygon  considered,  can  be  determined  by  means  of 
the  homologous  line  in  the  constructed  polygon. 


MEASUREMENT  OF  POLYGONS.        187 

PROPOSITION  VII.    THEOREM. 

267.  The  ratio  of  the  areas  of  two  similar  triangles 
is  equal  to  the  ratio  of  the  squares  of  their  homologous 
edges,  or  homologous  altitudes. 


B^ -sr*c  F 


Let  ABC  and  EFO  represent  two  similar  triangles, 
and  A  M  and  E  N  Iwmologous  altitudes. 


,L     *      A    B    C  B    C  AM 

To  brove  that        =-=  =          x   or  — 


EN2 

SUG.  1.  What  is  the  area  of  each  A?  What  is  the  ratio 
of  their  areas  in  terms  of  their  bases  and  altitudes  ? 

Sue.  2.  What  is  the  ratio  of  the  bases  in  terms  of  the 
altitudes  ? 

SUG.  3.  What,  then,  is  the  ratio  of  the  areas  of  the  As 
in  terms  of  their  bases  ?  What  in  terms  of  their  alti- 
tudes ? 

Therefore 


Ex.  159.  Draw  the  common  chord  of  two  intersecting 
circles.  From  any  point  in  the  chord  extended  draw 
tangents  to  the  two  circles.  Prove  that  the  tangents  are 
equal. 

Ex.  160.  To  construct  a  triangle  similar  to  a  given  tri- 
angle, having  a  given  perimeter. 


188 


PLANE    GEOMETRY. 


PROPOSITION  VIII    THEOREM. 

268.   The  ratio  of  the  areas  of  two  similar  polygons 
equals  the  ratio  of  the  squares  of  two  Jwmologous  sides. 


C' 


Let  O  and  O'  represent  two  similar  polygon* 

To  prove  that   -^  =  =^  . 

SUG.  1.  Divide  the  polygons  into  similar  As;  ABC 
similar  to  A'  B'  C' ,  etc. 

SUG.  2.  What  is  the  ratio  of  the  areas  of  As  A  B  C 
and  A'  B'  C'  in  terms  of  A  B  and  A'  B'  ?  What  is  the 
ratio  of  the  areas  of  As  A  CD  and  A'  C'  H  in  terms  of 
CD  and  C'  ZX  ?  Art.  267. 

SUG.  3.  Compare  the  ratios 
A  APE 


etc. 


O 


SUG.  4.  Compare  the  ratio  -=,-  with  either  of  the  ratios 

in  Sug.  3.     (Art.  222.) 

SUG.  5.  What  is  the  ratio  of  the  areas  of  the  polygons 
in  terms  of  the  sides  ?  (See  Sug.  2.) 

Therefore 

269.  COROLLARY.  The  ratio  of  the  areas  of  two  sim- 
ilar polygons  equals  the  ratio  of  the  squares  of  any  two 
homologous  lines. 


MEASUREMENT   OF    POLYGONS. 


PROPOSITION  IX.     THEOREM. 

270.  The  square  described  upon  the  hypotenuse  of 
a  right  triangle  is  equal  to  the  sum  of  the  squares  de- 
scribed upon  the  other  two  sides. 

N 


Let  ABC  represent  a  right  triangle,  whose  hypotenuse 
isAC,  and  let  A  E  be  the  square  upon  the  hypotenuse,  and 
B  S  and  B  Mthe  squares  on  the  other  two  sides. 

To  prove  that  A  E  =  B  S  plus  B  M. 

SUG.  1.  Draw  B  G  \\  to  C  E,  and  extend  to  meet  N  E, 
at  O.  Draw  B  N,  B  E,  C  S  and  A  M. 

SUG.  2.  Compare  As  B  C  E  and  A  CM. 

SuG.  3.  Compare  the  area  of  A  A  CM  with  the  area 
of  the  square  B  M,  and  the  area  of  A  B  C  E  with  the 
area  of  the  rectangle  C  O. 

SUG.  4.  Compare  the  area  of  the  square  B  M  with  the 
area  of  the  rectangle  C  O. 

SUG.  5.  Compare  As  C  A  S  and  NAB. 

SUG.  6.  Compare  the  square  B  S  with  the  rectangle  A  O. 

SUG.  7.  Compare  the  sum  of  the  areas  of  the  squares 
B  M  and  B  S,  with  the  sum  of  the  areas  of  the  rectangles 
C  O  and  A  O\i.  e.,  with  the  area  of  the  square  C  N. 

Therefore 


190  PLANE    GEOMETRY. 

271.  SCHOLIUM  I.     This  proposition  is  known  as  the 
Pythagorean  proposition.     It  is  so  named  in  honor  of 
Pythagoras,    who  is  supposed  to  have  given  the  first 
demonstration  of  it. 

272,  SCHOLIUM  II.     The  "square  of  a  line'*  is  an 
expression   often  used  instead  of  the  square  described 
upon  a  line.     But  if  the  former  expression  is  used  it  must 
be  interpreted  as  already  explained  in  Art.  260.     Using 
this  language,  the  Pythagorean  proposition  is  often  stated 
as  follows:   The  square  of  the  hypotenuse  of  a  right  tri- 
angle equals  the  sum  of  the  squares  of  the  other  two 
sides.     The  proposition  is  often  put  in  the  form  of  an 
equation,  thus:     A~C2   =  ^4~2?2   +   B~C* .      From  this 
equation  two  others  are  readily  obtained  by  mere  trans- 
position, viz.: 

A~B*  =^T2  -B~C* 

and  'B~C*  =  A~C*  -  ~AB'\ 

When  any  two  of  the  three  sides  of  a  right  triangle  are 
given,  the  remaining  side  may  be  found  from  one  of  the 
three  equations  just  written. 


Ex.  161.  If  two  triangles  have  an 
angle  of  one  equal  to  an  angle  of  the 
other,  the  ratio  of  their  areas  equals 
the  ratio  of  the  products  of  the  sides 
including  the  equal  angles. 

A  A  B  C         A  B  X  A  C 
°Ve   A  A  B'  C  ==  AB'xA~CT' 

Sue.   Connect   B  and    C'.      Compare  each   A    with 
A  A  B  C'  (Prop.  V,  Cor.). 


MEASUREMENT   OF    POLYGONS. 


191 


PROPOSITION  X.    THEOREM. 

273.  The  square  described  on  the  sum  of  two  lines 
equals  the  sum  of  the  squares  described  on  the  two 
lines  plus  twice  the  rectangle  whose  sides  are  the  two 
lines. 


M  /V 

Let  A  ~B  and  B  C  be  two  given  lines,  and  A  C  their  sum. 
Let  AN  be  tJie  square  described  on  A  C. 

To  prove  that  A  N  equals  the  sum  of  the  squares  de- 
scribed on  A  B  and  B  C  plus  twice  the  rectangle  whose  sides 
are  A  B  and  B  C. 

SUG.  1.  Lay  off  A  E  equal  to  A  B,  and  draw  E  O  ||  to 
A  C;  also  draw  B  M\\  to  C  N. 

SUG.  2.  What  kind  of  a  figure  is  E  Bl 

SUG.  3.  Compare  C  O  with  A  B  ';  also  O  N  and  M  N 
with  B  C. 

SUG.  4.  What  kind  of  a  figure  is  M  Ol 

SUG.  5.  Compare  the  rectangles  B  O  and  E  M. 

SUG.  6.  Notice  that  the  square  A  N,  is  divided  into 
four  parts,  and  then  note  the  answers  to  Sugs.  2,  4  and  5. 

Therefore  - 

274.  SCHOLIUM.  With  the  interpretation  given  in 
Art.  260  for  the  square  of  a  line  and  the  product  of  two 
lines,  this  proposition  may  be  expressed  thus: 

If  A  C  =  AB  +  B  C 

2  +  ~B~C*  +  2  A  B  X  B  C. 


then 


C*  = 


192  PLANE  GEOMETRY. 

EXERCISES. 

162.  Prove  that  the   area  of  a  rhombus  equals  one 
half  the  product  of  its  diagonals. 

163.  Given  the  area  of  a  trapezoid  equal  to  104  sq.  ft., 
the  altitude  equal  to  6  ft.  and  the  difference  between  the 
bases  equal  to  2  ft.     Find  the  two  bases  of  the  trapezoid. 

164.  A  B  is  a  chord  of  a  circle.     C  D  is  a  diameter 
perpendicular  to  A  B^  and  intersecting  A  B  at  E.     C  E 
is  5  in.   and  A  C  10  in.      Find  the  diameter  of  the 
circle. 

165.  Find  the  dimensions  of  a  rectangle  whose  perim- 
eter is  16  in.  and  whose  area  is  15  sq.  in. 

166.  Find  the  dimensions  of  a  rectangle  whose  perim- 
eter is  28  ft.  and  whose  diagonal  is  10  ft. 

167.  If  the  hypotenuse  of  a  right  triangle  is  15  ft.  and 
the  ratio  of  the  legs  is  f-,  what  is  the  area  ? 

168.  Prove  that  the  area  of  an  equilateral  triangle  con- 
structed on  the  hypotenuse  of  a  given  right  triangle,  is 
equal  to  the  sum  of  the  areas  of  the  equilateral  triangles 
constructed  on  the  other  two  sides  of  the  given  right  tri- 
angle. 

169.  If  one  acute  angle  of  a  right  triangle  is  60°,  prove 
that  the  area  of  the  equilateral  triangle  constructed  on 
the  hypotenuse  is  equal  to  the  area  of  a  rectangle  whose 
sides  are  the  two  legs  of  the  right  triangle. 

170.  The  altitude  of  a  trapezoid  is  3  ft.  and  the  bases 
are  8  and  12  ft.  respectively.     Extend  the  non-parallel 
sides  until  they  meet,  and  find  the  areas  of  the  two  tri- 
angles of  which  the  trapezoid  is  the  difference. 

171.  If  two  triangles  have  two  sides  of  one  equal  re- 
spectively  to   two  sides  of  another   and  the   included 
angles  supplementary,  the  triangles  are  equal  in  area. 


MEASUREMENT   OF    POLYGONS. 


193 


PROPOSITION  XI.     THEOREM. 

275.  The  square  described  on  the  difference  of  two 
lines  equals  the  sum  of  the  squares  described  on  the 
two  lines  minus  twice  the  rectangle  whose  sides  are 
the  two  lines. 


c 


G 


Let  A  "B  and  B  C  be  two  given  lines,  and  A  C  their  dif- 
ference. Let  A  F  be  the  square  described  on  A  C. 

To  prove  that  A  F  equals  the  sum  of  the  squares  de- 
scribed on  A  B  and  B  C,  minus  twice  the  rectangle  whose 
sides  are  A  B  and  B  C. 

STJG.  1.  Let  A  D  be  the  square  described  on  A  B  and 
B  K  the  square  described  on  B  C. 

SUG.  2.  Extend  K  C  to  Z,,  and  HFioG. 

SuG.  3.  Compare  E  B  with  B  C,  and  B  G  with  C  A. 
Compare  E  G  with  A  B. 

SuG.  4.  Compare  the  rectangle  E  F  with  the  rect- 
angle D  H. 

SUG.  5.  Notice  that  the  square  A  F  equals  the  sum  of 
the  squares  A  D  and  B  K>  minus  the  sum  of  the  rect- 
angles E  /^and  D  H. 

Therefore  - 


13  -Geo. 


194: 


PLANE    GEOMETRY. 


276.  SCHOLIUM.  With  the  interpretation  given  in 
Art.  260  for  the  square  of  a  line  and  the  product  of  two 
lines,  this  proposition  may  be  expressed  thus: 

If  AC=AB-BC 

then 


AC*  **AB*  +3  C'    —ZABx  BC. 


277.  The  projection  of  a  point  upon  a  straight  line 
is  the  foot  of  the  perpendicular  from  the  point  to  the  line. 

The  projection  of  a  given  straight 
line  upon  another  straight  line  is  that 
part  of  the  second  line  included  be- 
tween the  projections  of  the  extrem- 
ities of  the  given  line.  The  projection  £ 
ofAJB  upon  C  D  is  M  N. 


B 


M 


N 


PROPOSITION  XII.     THEOREM. 

278.  The  square  on  the  side  opposite  an  acute  angle 
of  a  triangle  equals  the  sum  of  the  squares  of  the  other 
two  sides  minus  twice  the  product  of  one  side  by  the 
projection  of  the  other  side  upon  that  side. 
G 


Let  ABCbea  triangle  of  wJiich  the  angle  A  is  acute, 
and  let  B  L,  A  D  and  A  K  1>e  the  squares  described  on  ttie 
sides  B  C,  AC  and  A  B  respectively. 


MEASUREMENT  OF  POLYGONS.        195 

To  prove  that  the  square  B  L  equals  the  sum  of  the 
squares  A  D  and  A  K,  minus  twice  A  B  times  the  projec- 
tion of  A  C  upon  A  B. 

SUG.  1.  From  A,  draw  A  O  _L  to  B  C,  and  extend  it 
to  R.  From  B,  draw  Bm  _L  to  A  C,  and  extend  it  to  E. 
From  C,  draw  Cn  _L  to  A  B  and  extend  it  to  H. 

SUG.  2.   Draw  A  /,  A  L,  B  D,  B  F,  C^and  C  G. 

SUG.  3.  Compare  A  A  B  /with  A  C  B  K.  Compare 
A  A  B  I  with  rectangle  B  R,  and  A  C  B  I  with  rect- 
angle B  H.  Compare  rectangle  B  H  with  rectangle  B  R. 

SUG.  4.  In  a  similar  manner  compare  rectangle  C  E 
with  rectangle  C  R. 

SUG.  5.  Compare  A  C  A  G  with  A  B  A  F.  Compare 
A  C  A  G  with  rectangle  A  H.  Compare  A  B  A  /'with 
rectangle  A  E.  Compare  the  rectangle  A  H  with  rect- 
angle A  E. 

SUG.  6.  The  square  B  L  =  the  sum  of  the  rectangles 
B  H  and  C  E.  Why  ? 

SUG.  7.  The  square  B  L  «=  the  sum  of  the  squares  A  K 
and  A  D,  minus  the  sum  of  what  two  rectangles  ? 

SUG.  8.  The  square  B  L  =  the  sum  of  the  squares  A  K 
and  A  D,  minus  twice  the  rectangle  A  H.  Why  ?  (That 
is,  the  square  B  L  =  the  sum  of  the  squares  A  K  and 
A  D,  minus  2  x  A  G  X  An.) 

SUG.  9.  Compare  A  G  X  An  with  A  B  X  An. 

SUG.  10.  Notice  that  An  is  the  projection  of  A  C 
upon  A  B. 

Therefore 


Ex.  172.  The  non-parallel  sides  of  a  trapezoid  are  each 
15  inches,  and  the  bases  are  12  inches  and  30  inches  re- 
spectively. Find  the  area  of  the  trapezoid. 


196 


PLANE    GEOMETRY. 


PROPOSITION  XIII.     THEOREM. 

279.  The  square.on  the  side  opposite  an  obtuse  angle 
of  a  triangle  equals  the  sum  of  the  squares  of  the  other 
two  sides  plus  twice  the  product  of  one  of  the  sides  "by 
the  projection  of  the  other  side  upon  that  side. 


Let  A  B  Cbea  triangle,  of  which  tlie  angle  A  is  obtuse; 
and  let  JB  L,  A  K  and  A  Dbe  squares  on  the  sides  JB  C,AB 
and  A  C  respectively. 

To  prove  thai  the  square  B  L  equals  the  sum  of  tht 
squares  A  K  and  A  D  plus  twice  A  B  times  the  projection 
of  A  C  upon  A  -B. 

SUG.  1.  From  A,  draw  A  O  J_  to  B  C,  and  extend  to  R. 
From  B,  draw  Bm  _L  to  C  A  extended,  and  extend  Bm 
to  E.  From  C,  draw  Cn  _L  to  B  A  extended,  and  extend 
Cn  to  H. 

SUG.  2.  Draw  A  7,  A  L,  B  D,  B  F,  CJ^smdC  G. 

SUG.  3.  Compare  As  A  B  I  and  C  B  K.  Compare 
A  A  B  I  with  rectangle  B  R,  and  A  C  B  K  with  rect- 
angle B  H.  Compare  rectangle  B  R  with  rectangle  B  H. 


MEASUREMENT  OF  POLYGONS.        197 

SUG.  4.  In  a  similar  manner  compare  rectangle  C  R 
with  rectangle  C  E. 

SUG.  5.  Compare  As  B  A  F  and  C  A  G.  Compare 
A  B  A  F  with  rectangle  A  E,  and  A  C  A  G  with  rect- 
angle A  H.  Compare  rectangle  A  E  with  rectangle  A  H. 

SUG.  6.  The  square  B  L  equals  the  sum  of  the  rect- 
angles B  H  and  C  E.  Why  ? 

SUG.  7.  The  square  B  L  equals  the  sum  of  the  squares 
A  K  and  A  D  plus  twice  the  rectangle  A  H.  Why  ? 
(That  is,  the  square  B  L  =  the  sum  of  the  squares  A  K 
and  A  D  plus  2  x  A  G  X  An.*) 

SUG.  8.   Compare  A  G  X  An  with  A  B  X  An. 

SUG.  9.  Notice  that  An  is  the  projection  of  A  C  upon 
A  B. 

Therefore 


Ex.  173.  If  one  circle  is  inscribed  in  a  right  triangle, 
and  another  circle  circumscribed  about  the  same  right 
triangle,  the  sum  of  the  diameters  of  the  circles  is  equal 
to  the  sum  of  the  two  legs  of  the  right  triangle. 

Ex.  174.  If  a  circle  is  inscribed  in  a  right  triangle  the 
sum  of  the  two  legs  of  the  triangle  exceeds  the  hypote- 
nuse by  an  amount  equal  to  the  diameter  of  the  circle. 

Ex.  175.  The  circle  inscribed  in  an  equilateral  triangle 
has  the  same  center  as  the  circle  circumscribed  about  the 
same  triangle,  and  the  radius  of  the  circumscribed  circle 
is  double  that  of  the  inscribed  circle. 

Ex.  176.  The  ratio  of  similitude  of  two  similar  poly- 
gons is  fy,  and  the  sum  of  their  areas  is  518  sq.  in.  Find 
the  area  of  each  polygon. 

Ex.  177.  Upon  a  given  base  construct  a  right  triangle 
having  given  the  perpendicular  from  the  right  angle  to 
the  hypotenuse. 


198  PLANE    GEOMETRY. 

PROPOSITION  XIV.     PROBLEM. 

280.   Upon  a  given  line  as  base,  to  construct  a  rect- 
angle equal  in  area  to  a  given  square. 


A 


Jjet  CD  be  the  given  line,  and  A  E  the  given  square. 

To  construct  upon  C  D  as  a  base,  a  rectangle  equal  in 
area  to  the  square  A  R. 

SUG.  1.  lyet  X  stand  for  the  required  altitude.  Suice 
the  area  of  the  rectangle  is  to  be  equal  to  the  area  of  the 
square,  the  product  of  the  measures  of  C  D  and  X  must 
equal  the  square  of  the  measure  of  A  B.  Hence, 
ABxAB=CDxX. 

SUG.  2.  Make  a  proportion  from  the  equation  in  Sug.  1. 
Art.  203. 

SUG.  3.  All  the  terms  of  the  proportion,  except  X,  be- 
ing given,  construct  X  by  a  method  based  upon  a  propo- 
sition involving  proportion. 

SuG.  4.  Having  now  constructed  X,  construct  the  rect- 
angle. 

Therefore 


Ex.  178.   Find  the  base  and  altitude  of  a  rectangle 
whose  diagonal  is  15  ft.  and  whose  area  is  108  sq.  ft. 


MEASUREMENT   OF    POLYGONS. 


199 


PROPOSITION  XV.     PROBLEM. 

281.   To  constmct  a  square  equal  in  area  to  a  given 
rectangle* 


B 


Let  A  B  C  D  be  a  given  rectangle* 

To  construct  a  square  equal  in  area  to  rectangle  A  C. 

SUG.  1.  What  is  the  area  of  A  C? 

SUG.  2.  If  X  represents  the  side  of  the  square  equal  in 
area  to  A  C  make  a  proportion  of  D  C,  B  C  and  X. 

SUG.  3.  Construct  the  line  represented  by  X  in  the 
proportion. 

Note . —  There  are  at  least  three  previous  propositions  by  which  the 
above  problem  can  be  worked.  It  may  be  time  well  used  by  the  pu 
pil  to  work  more  than  one  solution. 


Ex.  179.  Find  the  area  of  a  right  triangle  whose 
hypotenuse  is  1  ft.  8  in.,  and  one  of  whose  legs  is  1  ft. 

Ex.  180.  Construct  a  square  equal  in  area  to  a  given 
trapezoid. 

Ex.  181.  Construct  a  square  equal  in  area  to  a  given 
triangle. 

Ex.  182.  Construct  a  square  equal  in  area  to  a  given 
parallelogram. 


200 


PLANE    GEOMETRY. 


PROPOSITION  XVI.     PROBLEM. 

282.  To  construct  a  square  whose  area  is  equal  to 
the  sum  of  the  areas  of  two  given  squares. 


Let  A  and  U  be  two  given  squares. 

To  construct  a  square  whose  area  equals  the  sum  of  the 
areas  of  A  and  B. 

SUG.  Apply  Art.  270. 

PROPOSITION  XVII.     PROBLEM. 

283.  To  construct  a  square  whose  area  is  equal  to 
the  difference  of  the  areas  of  two  given  squares. 

SUG.  Apply  Art.  272. 

PROPOSITION  XVIII.     PROBLEM. 

284.  To  construct  a  rectangle  in  which  the  sum  of 
the  base  and  altitude  equals  a  given  linef  and  the  area 
equals  the  area  of  a  given  square. 

A  B 


F  E 

Let  C  D  be  the  given  line,  and  ABEF the  given squcvre. 


MEASUREMENT  OF  POLYGONS.        201 

To  construct  a  rectangle  in  which  the  sum  of  the  base  and 
altitude  equals  C  D,  and  the  area  equals  the  area  of  the 
square  A  B  E  F. 

SUG.  1.  The  side  A  B  of  the  given  square  must  be  a 
mean  proportional  between  the  two  segments  into  which 
C  D  must  be  divided. 

SUG.  2.  See  some  former  proposition  about  a  mean 
proportional:  for  example,  Art.  239,  II. 

QUERY.  Can  only  one,  or  can  an  indefinite  number,  of 
rectangles  be  constructed  whose  area  equals  the  area  of 
the  given  square  ? 


Ex.  183.  Prove  that  the  product  of  the  sum  of  two 
lines  by  their  difference  equals  the  difference  of  their 
squares. 

Ex.  184.  Construct  a  square  whose  area  is  equal  to  the 
sum  of  the  areas  of  three  given  squares. 

Ex.  185.  Upon  a  given  line  as  a  base,  construct  a  rect- 
angle whose  area  is  equal  to  the  sum  of  the  areas  of  a 
given  square,  a  given  trapezoid  and  a  given  triangle. 

Ex.  186.  Construct  a  parallelogram  having  a  given 
angle,  whose  base  and  altitude  taken  together  equal  a 
given  line,  and  whose  area  equals  the  area  of  a  given 
square. 

Ex.  187.  Two  parallel  chords  of  a  circle  are,  respect- 
ively, 36  in.  and  48  in.  long;  the  radius  of  the  circle  is 
30  in.  What  is  the  distance  between  the  chords  ? 

Ex.  188.  The  area  of  a  square  inscribed  in  a  circle  is 
one  half  the  area  of  a  square  circumscribed  about  the 
same  circle. 


202  PLANE    GEOMETRY. 

PROPOSITION  XIX.    THEOREM. 

285.  To  construct  a  triangle  equal  in  area  to  a  given 
polygon. 


Let  A  B  C  D  E  Fbea  given  polygon. 

To  construct  a  triangle  equal  in  area  to  the  polygon 
A  B  CD  EF. 

SuG.  1.  Draw  a  diagonal  cutting  off  one  vertex;  as  B  D. 

SUG.  2.  Through  C,  the  vertex  cut  off,  draw  a  line 
||  to  the  diagonal  B  D. 

SUG.  3.  Extend  E  D  until  it  meets  the  line  ||  to  the 
diagonal  at  O.  Connect  O  with  B. 

SUG.  4.   Compare  the  areas  of  the  As£CD  and  BOD. 

SUG.  5.  Compare  the  areas  of  the  polygons  A  B  CD  EF 
m&ABOEF. 

SUG.  6.  Notice  that  the  number  of  sides  of  the  last 
polygon  is  one  less  than  of  the  original  polygon. 

Continue  the  process  until  the  polygon  is  reduced  to  a 
triangle. 


Ex.  189.  The  area  of  a  triangle  is  equal  to  one  half 
the  product  of  its  perimeter  by  the  radius  of  the  inscribed 
circle. 


MEASUREMENT  OF  POLYGONS.        203 

PROPOSITION  XX.     PROBLEM. 

286.   To  find  two  straight  lines  which  have  the  same 
ratio  as  two  given  similar  polygons. 


Let  A  and  B  be  two  given  similar  polygons. 

To  find  two  lines  whose  ratio  equals  the  ratio    -^. 

SuG.  1.  Construct  a  right  triangle,  ABC,  having  for 
its  legs  A  B  and  B  C,  any  two  homologous  lines  in  the 
two  polygons,  and  from  B,  the  vertex  of  the  rt.  Z.,  drop 
a  _L  B  D  upon  the  hypotenuse  A  C. 

SuG.  2.  Compare  the  ratio  P°y£on        wjth  the  ratio 

polygon  N 

A    n  2  AD 

—  .     Compare  the  last  ratio  with  the  ratio  ^-^ . 
SuG.  3.  What,  then,  are  the  required  lines? 


Ex.  190.  If  tangents  to  a  circle  be  drawn  at  the  ex- 
tremities of  any  chord,  these  tangents  make,  with  each 
other,  an  angle  which  is  twice  the  angle  between  the 
chord  and  the  diameter  of  the  circle  drawn  through  the 
extremity  of  the  chord. 

Ex.  191.  Any  straight  line  drawn  through  the  center 
of  a  parallelogram  divides  the  parallelogram  into  two 
parts  which  are  equal  in  all  respects. 


204:  PLANE    GEOMETRY. 


PROPOSITION  XXI.     PROBLEM. 

287.  To  construct  a  polygon  similar  to  a  given  poly- 
gon and  equal  in  area  to  another  given  polygon. 


Let  A  and  B  be  two  given  polygons. 

To  construct  a  polygon  similar  to  A,  and  equal  in  area  to  B. 

SUG.  1.  Construct  squares  equal  in  area  to  the  polygons 
A  and  B  respectively,  and  let  m  and  n  represent  the 
sides  of  the  squares. 

SUG.  2.  L,et  a  represent  one  side  of  the  polygon  A, 
and  find  a  fourth  proportional  to  m>  n  and  a. 

SUG.  3.  Let  p  represent  the  fourth  proportional  just 
found,  and  construct  a  polygon  P  similar  to  A,  having 
the  side  p  in  the  polygon  P  homologous  to  the  side  a  in 
the  polygon  A. 

SUG.  4.  Compare  the  ratio  -p-  with  the  ratio  — j. 

A  2 

SUG.  5.  Compare  the  ratios  -^  and  — ^ .  Compare 
-—^  with  — 3- .  (See  Sugs.  2  and  3.)  Compare  the  ratios 
-=-  and  — 2- .  Compare  B  with  P. 

Jr 

288.  A  line  is  divided  into  mean  and  extreme  ratio 
when  the  ratio  of  the  whole  line  to  the  greater  segment 
equals  the  ratio  of  the  greater  to  the  lesser  segment. 


MEASUREMENT  OF  POLYGONS. 


205 


PROPOSITION  XXII.     PROBLEM. 

289.  To  divide  a  straight  line  into  mean  and  exr 
treme  ratio. 


Let  ABbea  given  straight  line. 

To  divide  A  B  into  mean  and  extreme  ratio. 

SUG.  1.  At  B,  erect  a  _L  to  A  B,  and  on  this  _L  lay  off 
B  O  =  \A  B. 

With  O  as  a  center,  and  O  B  as  a  radius,  describe  a 
circumference.  Draw  a  line  through  A  and  O,  meeting 
the  circumference  at  N  and  M. 

On  A  £  lay  off  a  distance  A  X  equal  to  A  N. 
A  M        A  B 


0 
.  2. 


AN 
SUG.  3.  By  division  -- 


-. 

Giveauth. 

BX 


_. 
Explain. 


SUG.  4.  Therefore,    by     inversion    and    substitution 
AB  _  AX 
AX~''  XB' 


PROPOSITIONS  IN  CHAPTER  IV. 


PROPOSITION  I. 

Two  parallelograms  having  equal  bases  and  equal  altitudes  are 
equal  in  area. 

PROPOSITION  II. 

If  two  rectangles  have  equal  altitudes,  the  ratio  of  their  areas  equals 
the  ratio  of  their  bases. 

PROPOSITION  III. 

The  number  of  units  of  area  in  any  rectangle  is  equal  to  the  prod- 
uct of  the  number  of  linear  units  in  the  base  and  altitude. 

PROPOSITION  IV. 

The  area  of  a  parallelogram  equals  the  product  of  its  base  and 
altitude. 

PROPOSITION  V. 

The  area  of  a  triangle  equals  one  half  the  product  of  its  base  and 
altitude. 

PROPOSITION  VI. 

The  area  of  a  trapezoid  equals  one  half  the  product  of  its  altitude 
by  the  sum  of  its  bases. 

PROPOSITION  VII. 

The  ratio  of  the  areas  of  two  similar  triangles  is  equal  to  the  ratio 
of  the  squares  of  their  homologous  sides,  or  homologous  altitudes. 

PROPOSITION  VIII. 

The  ratio  of  the  areas  of  two  similar  polygons  equals  the  ratio  of 
the  squares  of  two  homologous  sides. 

PROPOSITION  IX. 

The  squnre  described  upon  the  hypotenuse  of  a  right  triangle  is 
equal  to  the  sum  of  the  squares  described  upon  the  other  two  sides. 


MEASUREMENT    OF    POLYGONS.  207 


PROPOSITION  X. 

The  square  described  on  the  sum  of  two  lines  equals  the  sum  of 
the  squares  described  on  the  two  lines  plus  twice  the  rectangle  whose 
sides  are  the  two  lines. 

PROPOSITION  XI. 

The  squares  described  on  the  difference  of  two  lines  equals  the  sum 
of  the  squares  described  on  the  two  lines  minus  twice  the  rectangle 
whose  sides  are  the  two  lines. 

PROPOSITION  XII. 

The  square  on  the  side  opposite  an  acute  angle  of  a  triangle  equals 
the  sum  of  the  squares  oc  the  other  two  sides  minus  twice  the  product 
of  one  side  by  the  projection  of  the  other  side  upon  that  side. 

PROPOSITION  XIII. 

The  square  on  the  side  opposite  an  obtuse  angle  of  a  triangle  equals 
the  sum  of  the  squares  on  the  other  two  sides  plus  twice  the  product 
of  one  of  the  sides  by  the  projection  of  the  other  side  upon  that  side- 

PROPOSITION  XIV. 

Upon  a  given  line  as  base,  to  construct  a  rectangle  equal  in  area  to 
a  given  square. 

PROPOSITION  XV. 
To  construct  a  square  equal  in  area  to  a  given  rectangle. 

PROPOSITION  XVI. 

To  construct  a  square  whose  area  is  equal  to  the  sum  of  the  areas 
of  two  given  squares. 

PROPOSITION  XVII. 

To  construct  a  square  whose  area  is  equal  to  the  difference  of  the 
areas  of  two  given  squares. 

PROPOSITION  XVIII. 

To  construct  a  rectangle  in  which  the  sum  of  the  base  and  altitude 
equals  a  given  line  and  the  area  equals  the  area  of  a  given  square. 


208  PLANE    GEOMETRY. 

PROPOSITION  XIX. 
To  construct  a  triangle  equal  in  area  to  a  given  polygon. 

PROPOSITION  XX. 

To  find  two  straight  lines  which  have  the  same  ratio  as  two  given 
similar  polygons. 

PROPOSITION  XXI. 

To  construct  a  polygon  similar  to  a  given  polygon  and  equal  in 
area  to  another  given  polygon. 

PROPOSITION  XXII. 
To  divide  a  straight  line  into  mean  and  extreme  ratio. 


CHAPTER  V. 
REGULAR  POLYGONS  AND  CIRCLES. 


DEFINITION. 

290.  A  regular  polygon  is  a  polygon  which  is  both 
equilateral   and   equiangular.     The  equilateral  triangle 
and  square  are  examples  of  regular  polygons. 

PROPOSITION  I.    THEOREM. 

291.  An  equilateral  polygon  inscribed  in  a  circle  is 
a  regular  polygon. 


Let  ABC  D  EF  represent  an  equilateral  polygon  in- 
scribed in  a  circle. 

To  prove  that  A  B  C  D  E  F  is  a  regular  polygon. 

SUG.  1.  Compare  the  arcs  A  B,  B  C,  etc.     Give  auth. 
SUG.  2.  Compare  Zs  A  B  C,  B  C  D,  etc.     Give  auth. 
SUG.  3.  Apply  the  definition  of  a  regular  polygon  to 
A  B  CD  EF. 

Therefore 

14-Geo. 


210  PLANE    GEOMETRY. 

292.  COROI/LARY.     A  regular  polygon  may  have  any 
number  of  sides.     For,  if  the  circumference  of  a  circle  be 
divided  into  any  number  of  equal  parts,  the  lines  joining 
the  points  of  division  will  form  an  inscribed  equilateral 
polygon.     How  does  the  number  of  sides  of  the  potygon 
compare  with  the  number  of  parts  into  which  the  circum- 
ference is  divided  ? 

PROPOSITION  II.     THEOREM. 

293.  A.  circle  can  be  circumscribed  about  a  regular 
polygon,  and  a  circle  can  be  inscribed  in  a  regular 
polygon. 

A      M 


Let  A  B  C  D  E  F  represent  a  regular  polygon. 

I.  To  prove  that  a  circle  can  be  circumscribed  about 
A  B  CDEF. 

SUG.  1.  At  M  and  N,  the  middle  points  of  two  adja- 
cent sides,  erect  JLs  and  extend  them  until  they  meet 
at  O ;  then  O  is  the  center  of  a  circumference  which 
passes  through  A,  B  and  C.  Why  ? 

SUG.  2.  Join  O  with  each  vertex  of  the  polygon. 

SUG.  3.  Compare  O  C  with  O  B. 

SUG.  4.  Compare  Z  O  B  C  with  Z.  O  C  B. 

SUG.  5.   Compare  Z  A  B  C  with  Z  B  C  D. 

SUG.  6.  Compare  Z.  O  B  A  with  Z  O  CD. 


REGULAR    POLYGONS  —  CIRCLES.  211 

SUG.  7.  Compare  A  O  B  A  with  A  O  C  D,  and  hence, 
compare  O  A  with  O  D. 

SUG.  8.  The  circumference  through  A,  B  and  C,  also 
passes  through  D.  Why  ? 

SUG.  9.  In  a  similar  manner  show  that  the  same  cir- 
cumference passes  through  E  and  F. 

SUG.  10.  Can  a  O  be  circumscribed  about  the  polygon 
ABCDEFJ 

II.  To  prove  that  a  circle  can  be  inscribed  in  the  poly- 
gon A  B  C  D  E  F. 

SUG.  1.  Compare  the  distances  of  the  various  sides  of 
the  polygon  from  O.  (See  Art.  153.) 

SUG.  2.  Can  a  circle  be  inscribed  in  the  polygon 
ABCDEF1 

Therefore 

294.  The  radius  of  a  regular  polygon  is  the  radius 
of  the  circumscribed  circle. 

The  apothem  of  a  regular  polygon  is  the  radius  of 
the  inscribed  circle. 

The  center  of  a  regular  polygon  is  the  center  of  the 
inscribed  or  circumscribed  circle. 

The  angle  at  the  center  of  a  regular  polygon  is  the 
angle  formed  by  two  radii  drawn  to  two  adjacent  vertices 
of  the  polygon. 

295.  From  the  definitions  just  given  three  inferences 
may  be  immediately  drawn: 

(1)  The  angle  at  the  center  of  a  regular  polygon  equals 
four  right  angles  divided  by  the  number  of  sides  of  the 
polygon. 

(2)  The  radius  of  a  regular  polygon  bisects  the  angle 
of  the  polygon  to  which  it  is  drawn. 

(3)  The  angle  at  the  center  of  a  regular  polygon  is  the 
supplement  of  the  interior  angle  of  the  polygon. 


212  PLANE    GEOMETRY. 

PROPOSITION  III.     THEOREM. 

296.  If  a  circumference  is  divided  into  any  num- 
ber of  equal  parts,  the  tangents  drawn  through  the 
points  of  division  form  a  regular  circumscribed  poly- 
gon. 


C 
K 

Let  the  circumference  ABCDE  Foe  divided  into  equal 
parts,  AB,BC,  etc.9  and  at  the  points  A,  B,  etc.,  let  tan- 
gents be  drawn  forming  the  polygon  G  H  K  L  M  N. 

To  prove  that  G  H  K  L  M  N  is  a  regular  polygon. 

SuG.  1.  By  what  is  the  /_  G  measured  ?  By  what  is 
the  Z.  H  measured  ?  Compare  ^/s  G  and  H.  Compare 
^s  H  and  K,  etc. 

SUG.  2.  Compare  the  As  A  G  B  and  B  H  C.  Com- 
pare the  As  B  H  C  and  C  K  D,  etc. 

SUG.  3.  Compare  GH  and  HK\  also,  H  XT  and  KL, 
etc. 

SUG.  4.  Apply  the  definition  of  a  regular  polygon  to 
GHK  LM  N. 

Therefore 

297.  COROU,ARY  I.  If  the  vertices  of  a  regular  in- 
scribed polygon  be  connected  with  the 
middle  of  the  arcs  subtended  by  the 
sides,  a  regular  inscribed  polygon  of 
double  the  number  of  sides  is  formed. 
(See  Art.  145.) 


REGULAR    POLYGONS  —  CIRCLES.  213 

298.  COROLLARY  II.     The  perimeter  of  a  regular  in- 
scribed polygon  is  less  than  the  perimeter  of  a  regular 
inscribed  polygon  of  double  the  number  of  sides. 

299.  COROLLARY  III.     If   a  regular 
polygon  is  circumscribed  about  a  circle, 
and  if  tangents  are  drawn  at  the  middle 
points  of  the  arcs  between  the  original 
points  of  contact,  a  regular  circumscribed 
polygon  of  double  the  number  of  sides  is 
formed. 

300.  COROLLARY  IV.     The  perimeter  of  a  regular  cir- 
cumscribed polygon  is  greater  than  that  of  a  regular  cir- 
cumscribed polygon  of  double  the  number  of  sides. 

301.  COROLLARY  V.     If  a  regular 
polygon  is  inscribed  in  a  circle,  tangents 
drawn  at  the  middle  points  of  the  arcs 
subtended  by  the  sides  of  the  inscribed 
polygon,  form  a  regular   circumscribed 
polygon  whose  sides  are  parallel  to  the 

sides  of  the  inscribed  polygon,  and  whose  vertices  lie  in 
the  radii  extended  of  the  inscribed  polygon. 

SUG.  1.  Draw  radii  of  the  O  to  the  points  of  contact  of 
tangents;  also  radii  to  the  vertices  of  the  inscribed  poly- 
gon, and  extend  these  latter  radii. 

SUG.  2.  Show  that  each  radius  of  the  latter  set  bisects 
the  Z  formed  by  two  adjacent  radii  of  the  former  set. 


Ex.  192.  The  line  joining  the  middle  points  of  the 
bases  of  a  trapezoid  divides  the  trapezoid  into  two  parts 
equal  in  area. 


214  PLANE    GEOMETRY. 

PROPOSITION  IV.     THEOREM. 

302.  Regular  polygons  of  the  same  number  of  sides 
are  similar 


H        K 


Let  A  J>  and  HP  represent  two  regular  polygons  of  the 
same  number  of  sides. 

To  prove  thai  A  D  and  HP  are  similar  polygons. 

SUG.  1.  Compare  the  sum  of  the  interior  Z!s  of  A  D 
with  the  sum  of  the  interior  ^s  of  HP. 

SUG.  2.  Compare  Zs  A  and  H,  B  and  K,  etc. 

SuG..  3.  Compare  the  ratios  -  —^  and  -=— - ,  etc. 

Jj   C  K.  L, 

A  B  B  C* 

SuG.  4.  Compare  the  ratios  -77-^=  and  -==-j  ,  etc. 

JTl    4\. 

Therefore 


Ex.  193.  Given  the  three  angles  of  a  triangle,  and  the 
area  equal  to  the  area  of  a  given  polygon;  construct  the 
triangle. 

SUG.  Construct  any  triangle  having  the  three  given 
angles,  and  then  consult  Art.  287. 

Bx.  194.  What  is  the  locus  of  the  center  of  a  circle 
which  is  tangent  to  a  given  line  at  a  given  point  ? 


REGULAR   POLYGONS — CIRCLES.  215 

PROPOSITION  V.    THEOREM. 

303,  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  have  the  same  ratio  as  their 
radii,  or  their  apothems. 


H  N  K 


Let  ABC  I>  E  F  and  HK  LPRT  represent  two  regu- 
lar polyyoiitt  of  the  same  number  of  sides. 

„    .  AB+ BC+CD+DE+EF+FA 

To  prove  that     HK+KL  +LP+PR  +R  T~+Ttf 
O  A          OM 
~~  Stf  ==   SJV' 

SUG.  1.  Compare  the  ratio  of  the  perimeters  with  the 

A  B 

ratio  of  any  two  homologous  sides;  as   -77^- 


SUG.  2.  Compare   ^j^  with    -^~F>,    or  with  -=-T>. 


-T> 

IV 

O  A 

STJG.  3.  Compare  the  ratio  of  the  perimeters  with  -prr>, 

o  ft 

.  ,     OM 
orwlth   __r. 

Therefore  - 

304.  COROLLARY.  The  areas  of  two  regular  polygons 
of  the  same  number  of  sides  have  the  same  ratio  as  the 
squares  of  their  radii,  or  the  squares  of  their  apothems. 


216  PLANE   GEOMETRY. 

PROPOSITION  VI.     THEOREM. 

305.  If  the  number  of  sides  of  a  regular  inscribed 
polygon  be  increased  indefinitely  the  apothem  is  a 
variable  which  approaches  the  radius  as  a  limit. 


Let  AB  be  a  side  of  a  regular  polygon,  and  O  H  its 
apothem. 

To  prove  that  if  the  number  of  sides  of  the  polygon  be 
increased  indefinitely,  the  apothem  is  a  variable  which  ap- 
proaches O  A  as  a  limit. 

SuG.  1.  In  the  A  A  O  H,  compare  O  A  and  O  H. 

SuG.  2.  Compare  O  A  —  O  H  with  A  H\  with  A  B. 

SuG.  3.  Since,  by  sufficiently  increasing  the  number 
of  sides  of  the  polygon,  one  side,  A  B,  may  be  made 
shorter  than  any  assumed  line,  however  short,  what  re- 
lation does  O  A  sustain  to  O  HJ  That  is,  what  relation 
does  the  radius  sustain  to  the  apothem  ? 

Therefore 


Ex.  195.  If  the  middle  points  of  two  adjacent  sides  of 
a  parallelogram  be  joined,  a  triangle  is  formed,  equal  in 
area  to  one  eighth  of  the  area  of  the  parallelogram. 


REGULAR  POLYGONS  —  CIRCLES.  217 

EXERCISES. 

196.  Draw  a  circle  through  a  given  point  tangent  to  a 
given  line  at  a  given  point. 

197.  The  diagonals  of  a  parallelogram  divide  it  into 
four  triangles  equal  in  area. 

198.  What  is  the  locus  of  the  vertex  of  an  isosceles  tri- 
angle upon  a  given  base  ? 

199.  A  B  C  is  a  triangle,  and  D  any  point  in  B  C  ex- 
tended.    Find  a  point  E  in  A  B,  such  that  the  area  of 
the  triangle  E  B  D  will  be  equal  to  the  area  of  the  tri- 
angle ABC. 

200.  Draw  a  line  through  a  given  point  in  a  side  of  a 
triangle  so  as  to  divide  the  triangle  into  two  parts  equal 
in  area. 

201.  If  the  diagonals  of  a  quadrilateral  intersect  at 
right  angles,  show  that  the  area  of  the  quadrilateral  is 
one  half  the  area  of  a  rectangle  whose  sides  are  equal  to 
the  diagonals  of  the  quadrilateral. 

202.  The  hypotenuses  of  three  isosceles  right  triangles 
form  a  right  triangle.     Prove  that  one  of  the  isosceles 
triangles  is  equal  to  the  sum  of  the  other  two. 

203.  If  the  diagonals  of  a  quadrilateral  intersect  at 
right  angles,  prove  that  the  sum  of  the  squares  on  one 
pair  of  opposite  sides  equals  the  sum  of  t-ie  squ'ares  on 
the  other  pair. 

204.  If  a  parallelogram  be  inscribed  in,   or  circum- 
scribed about,  a  circle,  the  diagonals  pass  through  the 
center. 

205.  A  circle  circumscribes  an  isosceles  triangle,  and 
tangents  are  drawn  to  the  circle  through  the  vertices  of 
the  triangle.     Prove  that  these  tangents  form  a  second 
isosceles  triangle,  and  that  the  two  triangles  cannot  have 
equal  vertical  angles  unless  both  are  equilateral. 


218  PLANE    GEOMETRY. 


PROPOSITION  VII.     THEOREM. 

306.  If  the  number  of  sides  of  a  regular  inscribed 
polygon  be  increased  indefinitely,  the  perimeter  of  the 
polygon  is  a  variable  which  approaches  the  circum- 
ference of  the  circle  as  a  limit. 


Let  CD  represent  the  side  of  a  regular  inscribed  poly- 
gon, and  p  its  perimeter. 

To  prove  that  as  the  number  of  the  sides  of  the  polygon 
increases^  p  is  a  variable  which  approaches  the  circumference 
of  the  circle  as  a  limit. 

SUG.  1.  L,et  E  B  be  the  side  of  a  circumscribed  poly- 
gon similar  to  the  inscribed  polygon,  and  let  P  denote 
the  perimeter  of  the  circumscribed  polygon.  L,et  r  de- 
note the  radius  and  a  the  apothem  of  the  inscribed  poly- 
gon. 

P         r 

SUG.  2.  Compare  the  ratios  -  and  — . 

f_          P          a 
SUG.  3.  The  ratio  — — ^  =  what?     See  Ex.  128. 

SUG.  4.  From  the  answer  to  Sug.  3,  find  the  value 


SUG.  5.  Since  P  is  greater,  and  p  less,  than  the  cir- 
cumference of  the  0,  the  difference  P  —  p  must  be 
greater  than  the  difference  between  the  circumference  and 
either  P  or/.  I^et  c  represent  the  circumference  of  the  O, 
then  c  —  p  is  less  than  P  —  p. 


REGULAR    POLYGONS  —  CIRCLES.  219 


SuG.  6.  Show  that  by  increasing  the  number  of  sides 
of  the  polygon,  the  value  of  P  —  p  (Sug.  4),  may  be 
made  to  decrease  indefinitely,  yet  cannot  be  made  equal 
to  zero. 

SUG.  7.  Hence,  as  c  —  p  is  less  than  P '  —  p  and  still 
c  —  p  cannot  be  made  equal  to  zero,  therefore,  c  is  the 
limit  of  p.  That  is,  the  circumference  of  the  O  is  the 
limit  of  the  perimeter  of  the  inscribed  polygon  when  the 
number  of  sides  of  the  polygon  is  made  to  increase  in- 
definitely. 

Therefore 

ADDITIONAL  HELPS. 

The  numbers  attached  to  the  following  additional 
helps  indicate  that  they  are  answers  to  suggestions  of 
corresponding  numbers. 

'•  >-.-• 

8.  fji---^. 

4.     P-p=(r~a)  x  -r. 

p 

6.      -  continually  decreases,  since  P  decreases  and  r 

remains  unchanged;  also,  r  —  a  diminishes  indefinitely 

p 

(Prop.  VI).     Hence    (r  —  a)    X  —    diminishes   indefin- 
itely, but  cannot  be  made  absolutely  zero. 


Ex.  206.  To  draw  a  common  tangent  to  two  given 
circles. 

Ex.  207.  If  three  equal  circles  are  tangents  to  one  an- 
other, the  lines  joining  their  centers  form  an  equilateral 
triangle. 


220 


PLANE    GEOMETRY. 


PROPOSITION  VIII.    THEOREM. 

307.  If  tine  number  of  sides  of  a  regular  inscribed 
polygon  be  increased  indefinitely,  the  area  of  the  poly- 
gon is  a  variable  which  approaches  the  area  of  the 
circle  as  a  limit. 


Let  C  J>  represent  a  side  of  a  regular  inscribed  polygon, 
and  let  A  denote  the  wrea  of  the  inscribed  polygon  and  A' 
the  area  of  the  circle. 

To  prove  that  if  the  number  of  sides  of  the  polygon  be  in- 
definitely increased,  A  is  a  variable  whose  limit  is  A'. 

SUG.  1.  Let  r  denote  the  radius,  and  a  the  apothem,  of 
the  inscribed  polygon;  also,  let/>  denote  the  perimeter  of 
the  inscribed  polygon,  and  P  the  perimeter  of  a  similar 
circumscribed  polygon.  Let  E  B  be  a  side  of  the  cir- 
cumscribed polygon  homologous  to  the  side  C  D  of  the 
inscribed  polygon. 

SUG.  2.  What  is  the  area  of  the  trapezoid  E  B  C  Dm 
terms  of  E  B,  CD  and  F  G  ?  That  is,  in  terms  of  E  B> 
CZ>and  (r  —  a). 

SUG.  3.  What  is  the  difference  of  the  areas  of  the  cir- 
cumscribed and  inscribed  polygons  in  terms  of  P,  p 
and  (r  —  a)  ? 


REGULAR   POLYGONS  —  CIRCLES.  221 

SUG.  4.  Let  A"  represent  the  area  of  the  circumscribed 
polygon.  Then  A"  —  A  equals  what  in  terms  of  P.  f> 
and  (r  —  a)  ? 

SUG.  5.   Compare  the  answer  to  Sug.  4  with  Px  (r—a). 

SUG.  6.  As  the  number  of  sides  of  the  polygon  is  in- 
creased, P  evidently  diminishes,  and  (r  —  a)  approaches 
zero  as  a  limit.  Therefore,  P  X  (r  —  a)  approaches  zero 
as  a  limit.  What,  then,  may  be  said  of  the  difference 
A"  —  A  as  the  number  of  sides  of  the  polygons  is  in- 
creased ? 

Therefore 

ADDITIONAL  HELPS. 

The  numbers  attached  to  the  following  additional  helps 
indicate  that  they  are  answers  to  suggestions  of  corre- 
sponding numbers. 

2.  Trapezoid  ABCD=(EB+  CD)X±  (r-  a). 
Why? 

4.  A"— A  =  (P+fi)x±(r—a).     Why? 

5.  (P+/)X£(r  — «)</>X(r--«).     Why? 

6.     A"  —  A  continually  decreases  and  approaches  zera 

as  a  limit. 

308.  COROLLARY.  The  area  of  the  circle  is  the  limit 
of  the  area  of  a  regular  circumscribed  polygon  as  the 
number  of  sides  is  indefinitely  increased. 


Ex.  208.  If  the  radius  of  one  circle  is  the  diameter  of 
another,  the  circles  are  tangent  to  each  other,  and  any 
line  drawn  from  the  point  of  contact  to  the  outer  circum- 
ference is  bisected  by  the  inner  one. 

Ex.  209.  Draw  a  circle  of  a  given  radius,  through  a 
given  point,  tangent  to  a  given  circle.  When  will  there 
be  two  solutions,  when  one  solution  and  when  no  solu- 
tion. 


222 


PLANE    GEOMETRY. 


PROPOSITION  IX.     THKOREM. 

309.   The  area  of  a  regular  polygon  is  equal  to  one 
half  the  product  of  its  perimeter  and  apothem. 


-D 


Let  AB  C  D  EF  be  a  regular  polygon,  and  O  N  its 
apothem. 

To  prove  that  the  area  of  the  polygon  equals  one  half  the 
product  of  its  perimeter  and  apothem.  « 

SUG.  1.  Circumscribe  a  circle,  and  draw  radii  to  the 
vertices  of  the  polygon. 

SuG.  2.  Compare  the  distances  from  the  center  to  the 
sides  of  the  polygon. 

SuG.  3.  The  area  of  the  A  A  O  B  equals  what  ?  Give 
an  expression  for  the  area  of  each  of  the  other  As  into 
which  the  polygon  is  divided.  The  sum  of  the  areas  of 
all  the  As  equals  what? 

Therefore 


Ex.  210.  Draw  a  circle  with  its  center  at  a  given  point 
tangent  to  a  given  circle.  When  is  there  only  one  solu- 
tion? 


REGULAR  POLYGONS  —  CIRCLES.  223 

EXERCISES. 

211.  Construct  a  triangle  equal  in  area  to  a  given  tri- 
angle, two  sides  of  tne  required  triangle  being  given. 
Show  when  there  are  two  solutions,  when  one  solution 
and  when  no  solution. 

212.  Construct  a  line  that  shall  equal  1^3. 

SUG.  Let  x  =  1/3".     Then  x*  =  3;    or,  -  =  ^ .    See 

X  JL 

Art.  249. 

213.  Construct  a  line  that  shall  equal  l/S  in.  Give  two 
solutions. 

214.  Construct  three  equal  circles  which  shall  be  tan- 
gent to  one  another  and  to  a  given  circle.     Consider  two 
cases:  (1)  that  in  which  the  three  circles  are  within  the 
given  circle,  and  (2)  that  in  which  the  three  circles  are 
without  the  given  circle. 

215.  If  the  diameter  of  a  circle  be  divided  into  two 
parts,  and  on   these  parts,  as  diameters,  circles  be  de- 
scribed, the  sum  of  the  circumferences  of  the  two  circles 
equals  the  circumference  of  the  given  circle. 

216.  Find  the  side  of  a  square  equal  in  area  to  a  rect- 
angle whose  sides  are  6  and  9. 

217.  Find  the  length  of  the  chord  joining  the  points  of 
contact  of  two  tangents  to  a  circle  whose  radius  is  8  in. 
drawn  from  a  point  12  in.  from  the  center. 

218.  Construct  a  triangle  whose  area  is  equal  to  9  times 
the  area  of  a  triangle  whose  sides  are  6,  7  and  9. 

219.  What  is  the  ratio  of  the  areas  of  two  similar  tri- 
angles whose  homologous  sides  have  the  ratio  f  . 

220.  Give.i  two  similar  triangles,  construct  a  third  tri- 
angle similar  to  the  other  two  whose  area  shall  be  equal 
to  the  sum  of  the  areas  of  the  other  two. 


224  PLANE    GEOMETRY. 


PROPOSITION  X .     THEOREM. 

310.  The  circumferences  of  two  circles  have  the 
same  ratio  as  their  radii. 


Let  O  and  S  represent  two  circles,  JB  and  H'  their  radii 
and  C  and  C'  their  circumferences. 

To  prove  that  ^  =  -~. 

SUG.  1.  Inscribe  in  the  two  circles  similar  regular  poly- 
gons. I,et  P  and  P'  represent  their  perimeters.  Then 

-p  =  -^7-  .     Give  auth. 

r> 

SUG.  2.  From  the  equation  in  Sug.  1,  P  =•  Px-=- . 

K 

SuG.  3.  Now,  let  the  number  of  sides  of  each  polygon 
continually  increase,  always,  however,  keeping  the  num- 
ber of  sides  of  one  polygon  the  same  as  the  number  of 
sides  of  the  other  polygon.  During  this  change  the 

r> 

variable  P  is  always  equal  to  the  variable  P  x  -^ .   Why? 

K 

SuG.  4.  Since  P  and  P  are  variables  which  approach 
C  and  C'  respectively,  as  their  limits  (Prop.  VII),  there- 

r> 

fore  P  and  P  X  -^-  are  variables  which  approach  C  and 

TO 

C'x-^7-  respectively,  as  their  limits.     Why  ? 


REGULAR    POLYGONS  —  CIRCLES.  225 


SUG.  5.   C=  C'xr.     Why? 
/t 

SUG.  6.   Hence    ^  =  ~  .     Why  ? 
Therefore  - 

311.  COROU,ARY  I.     The  circumferences  of  two  circles 
have  the  same  ratio  as  their  diameters. 

If  D  and  D'  represent  their  diameters, 
C        D_ 
C  ~=  £>'  ' 

312.  CORONARY  II.     The  ratio  of  the  circumference 
of  a  circle  to  its  diameter  is  the  same  for  all  circles. 

For,  by  Cor.  1,   ^  =  -jy  , 

C         C' 

and  by  alternation    -~  =   —  . 

This  ratio  is  represented  by  the  Greek  letter  n  (pi). 
Hence,  ^  =  n ;  or,  ^  =  n . 

Hence,  C  =  n  D ;  or,  C  =  2  n  R. 


Ex.  221.  Draw  a  line  through  the  vertex  of  a  triangle 
so  as  to  divide  the  triangle  into  parts  whose  areas  have 
the  ratio  f. 

Ex.  222.  Two  circles  are  tangent  externally;  locate  a 
line  of  a  given  length  so  that  it  shall  pass  through  the 
point  of  contact  and  have  its  extremities  in  the  circum- 
ferences of  the  circles. 

Ex.  223.  Given  one  leg  of  a  right  triangle  and  the  ra- 
dius of  the  inscribed  circle,  construct  the  triangle. 

Ex.  224.  Inscribed  a  circle  in  a  given  rhombus. 

15— Geo. 


226  PLANE    GEOMETRY. 

PROPOSITION  XI.     THEOREM. 

.    313,   The  areas  of  two  circles  have  the  same  ratio  as 
the  squares  of  the  radii. 


Let  A  and  A'  represent  the  areas,  and  M  and  H'  the 
radii  of  the  circles  O  and  O'  respectively. 

To  prove  that    -*r  =*   -fsr  • 
A  J\. 

SUG.  1.  Inscribe  in  the  two  circles  regular  polygons  of 
the  same  number  of  sides,  and  let  M  and  M1  represent 

their  respective  areas. 

M 
SUG.  2.  What  does  the  ratio    -^  equal  in  terms  of  R 

M'     u 

and  R  ?     (Art.  268.) 
SUG.  3.  Proceed  as  in  Prop.  X. 
Therefore  • 

314.  COROLLARY.     The  areas  of  circles  have  the  same 
ratio  as  the  squares  of  their  diameters. 


Ex.  225.  Construct  a  triangle  equal  in  area  to  a  given 
triangle,  the  base  and  the  distance  from  the  vertex  to 
the  middle  point  of  the  base  being  given. 


REGULAR    POLYGONS  —  CIRCLES.  227 

PROPOSITION  XII.    THEOREM. 

315.   The  area  of  a  circle  equals  one  half  the  product 
of  its  circumference  and  radius. 


Let  O  represent  a  circle,  H  its  radius,  C  its  circumfer- 
ence and  A  its  area. 

To  prove  that  A  equals  one  half  of  C  X  R. 

SUG.  1.  Inscribe  in  the  circle  a  regular  polygon,  and 
let  P  denote  its  perimeter,  Afits  area,  and  r  its  apothem. 

SUG.  2    Express  M  in  terms  of  P  and  r.     Give  auth. 

SUG.  3.  If  the  number  of  sides  of  the  polygon  be  in- 
definitely increased,  express  the  value  of  M. 

SUG.  4.  Express  the  limit  of  Mt  the  area  of  the  vary- 
ing polygon. 

SUG.  5.  Express  the  limit  of  the  varying  value  of  M. 

SUG.  6.   What  is  the  area  of  the  O  equal  to  ?     Why  ? 

Therefore 

ADDITIONAL  HELPS. 

The  numbers  attached  to  the  following  helps  indicate 
that  they  are  answers  to  the  above  suggestions  of  the 
same  numbers. 

2.     Af=$Pxr. 

3.     M  =•  \  Px  r  however  great  the  number  of  sides. 
4."    Limit  of  M  =  A.     (Art.  307.) 

5.  Since  the  limit  of  P  is  C  (Art.  306),  the  limit  of 
\P*R  is  \  CxR. 

6.  Therefore         A  =  %CxR.     (Art.  170.) 


228  PLANE    GEOMETRY. 

"*  316.  COROLLARY  I.  The  area  of  a  circle  equals  n  times 
the  square  of  the  radius.  If  the  value  of  C  given  in 
article  312  be  substituted  in  the  expression  for  the  area, 
the  above  equation  becomes 

A  =  \x2nRxR>, 
that  is,  A  =  TT  R*. 

317.  COROLLARY  II.     The  area  of  a  sector  equals  one 
half  the  product  of  radius  and  arc. 

SUG.  Compare  the  ratio  of  the  sector  to  the  circle,  with 
the  ratio  of  the  arc  of  the  sector  to  the  circumference  of 
the  circle. 

PROPOSITION  XIII.     THEOREM. 

318.  One  side  of  a  regular  hexagon  equals  the  ra- 
dius of  the  circumscribed  circle. 


Let  ABC  represent  a  regular  hexagon  inscribed  in  a 
cvrcle  whose  radius  is  O  A. 

To  prove  A  B  =  O  A.     . 

SUG.  1.  Draw  the  radius  O  B. 

SUG.  2.  The  Z.  A  O  B  is  what  part  of  four  rt.  Zs  ? 
Then,  how  many  degrees  are  there  in  Z.  A  O  B1 

SuG.  3.  How  many  degrees  in  the  sum  of  the  Z!s  O  A  B 
and  O  B  A  ? 

SUG.  4.  Compare  Zs  O  A  B  and  O  B  A.  How  many 
degrees  in  each  of  these 


REGULAR   POLYGONS  —  CIRCLES.  229 

SUG.  5.   Compare  the  three  Zs  of  the  A  A  O  B. 
SUG.  6.  Compare  the  sides  O  A  and  A  B. 
Therefore 

319.  Similar  arcs,  similar  sectors  and  similar  seg- 
ments in  different  circles  are  those  which  correspond  to 
equal  angles  at  the  center. 

PROPOSITION  XIV.     THEOREM. 

320.  The  areas  of  tivo  similar  segments  have  tlie 
same  ratio  us  the  squares  of  their  radii. 


Let  BCD  nnd  F  HK  represent  two  similar  segments 
of  circles  tvhose  centers  are  A  and  E  respectively. 


seg. 

SUG.  1.   Provethatthe  As^  #C  and  £^^7  are  similar. 

SUG.  2.  Express  the  ratio  of  the  areas  of  As  A  B  C  and 
E  F  H  in  terms  of  A  B  and  EF\  also  the  ratio  of  the  areas 
of  sectors  A  B  D  C  and  EFKH  in  terms  of  A  B  and  E  F. 

SUG.  3.  Compare  the  ratio  of  the  areas  of  the  sectors 
with  the  ratio  of  the  areas  of  the  As. 

SUG.  4.  Take  the  proportion  of  Sug.  3,  by  alternation 
and  then  by  division.  Compare  the  ratio  of  the  areas  of 
segments  with  the  ratio  of  the  areas  of  As  or  of  sectors. 


SUG.  5.  Compare  the  ratios  ^.  and 

seg.FHK 

Therefore  - 


230 


PLANE    GEOMETRY. 


PROPOSITION  XV.     PROBLEM. 

321.  Given  the  radius  of  a  circle  and  the  side  of  a 
regular  inscribed  polygon,  required  to  find  the  side  of 
a  regular  inscribed  polygon  of  double  the  number  of 
sides. 

C 


Let  O  represent  a  circle,  H  its  radius,  A  B  the  side  of  a 
regular  inscribed  polygon,  and  A  C  the  side  of  a  regular 
inscribed  polygon  of  double  the  number  of  sides. 

To  find  the  length  of  A  C  in  terms  of  the  known  quanti- 
ties, R  and  A  B. 

SUG.  1.  Connect  O  and  A. 

SuG.  2.  In  the  rt.  A  A  S  O,  express  the  value  of  5  O 
in  terms  of  O  A  (=  R}  andAS(=$A  B}.  (Art.  270.) 

SUG.  3.  Express  C  S  in  terms  of  R  and  £  O,  and  then 
in  terms  of  R  and  A  B. 

SUG.  4.  In  the  rt.  A  A  S  C,  A  S  and  C  S  have  both 
been  expresssed  in  terms  of  R  and  A  B.  Hence,  express 
A  C  in  terms  of  R  and  A  B,  and  reduce  the  answer  to  its 
simplest  form.  The  result  is: 


A  c  =  V' 

If  R  equals  unity 
A  C  = 


—  AB 


*-V±-A~B*. 


REGULAR   POLYGONS  —  CIRCLES.  231 

EXERCISES. 

226.  If,  from  a  point  without  a  circle,  two  secants  be 
drawn  whose  external  segments  are  8  in.  and  3  in.,  while 
the  internal  segment  of  the  latter  is  17  in.,  what  is  the 
internal  segment  of  the  former  ? 

227.  The  sides  of  a  triangle  are  5,  6  and  7,  and  the 
side  corresponding  to  6,  in  a  similar  triangle,  is  36;  find 
the  other  two  sides  of  the  triangle. 

228.  Find  the  altitude  of  an  equilateral  triangle  whose 
side  is  10  in.     Find  the  side  when  the  altitude  is  10  in. 

229.  If  three  -similar  polygons  be  constructed  on  the 
three  sides  of  a  right  triangle,  prove  that  the  area  of  the 
polygon  constructed  on  the  hypotenuse  equals  the  sum 
of  the  areas  of  the  polygons  constructed  on  the  other  two 
sides. 

SUG.  See  Art.  268. 

230.  All  equal  chords  of  any  circle  are  tangents  to 
some  other  circle. 

231.  In  A  B,  the  diameter  of  a  circle,  or,  in  A  B  ex- 
tended, take  any  point  C,  and  draw  CD  perpendicular 
to  A  B  ;  if  A  be  joined  with  any  point  P  in  CD,  and  A  P 
meet  the  circumference  at  Q\  then  A  PxA  Q=*ACx  AB. 
That  is,  A  P  X  A  Q  =  a  constant. 

232.  If  A  is  a  given  point,  and  P  any  point  in  a  given 
straight  line,  and  if  a  point  Q  be  taken  in  the  line  joining 
A  and  P,  so  that  A  P  x  A  Q  is  constant,  then,  as  Amoves 
along  the  given  line,  Q  will  move  on  the  circumference 
of  a  circle  which  passes  through  A.     (See  Ex.  231.) 

233.  If  A  B  be  the  diameter  of  a  circle,  and  if  a  point 
P  be  taken  on  any  chord  A  Q,  or  A  Q  extended,  so  that 
A  PxA  Q  is  constant,  the  locus  of  P  will  be  a  straight 
line  perpendicular  to  A  B.     (See  Ex.  231.) 


232 


PLANE    GEOMETRY. 


PROPOSITION  XVI.     PROBLEM. 

322.   To  compute,  approximately,  the  ratio  of  the 
circumference  of  a  circle  to  its  diameter. 
Consider  the  circle  whose  radius  is  unity. 


By  Prop.  XV,  A  C  =  J  2  -  1/4  -A~£*. 

If  A  Cis  one  of  the  equal  sides  of  a  regular  polygon  of 
n  sides,  n  X  A  C  =  the  perimeter,  which  may  be  re- 
garded as  an  approximation  to  the  circumference  of  the 
circle;  the  greater  the  number  of  sides  the  closer  the  ap- 
proximation. 

If  A  B  equals  one  side  of  a  regular  inscribed  hexagon, 
A  B  =  the  radius,  which  =  1. 

The  formula  then  gives 


-l/~3  =  .51763809. 


A  C=  A/2--  1/3T-  (I)2    = 

Hence,  the  perimeter  of  a  regular  twelve-sided  poly- 
gon =  12  x.  51763809  =  6.21165708. 

The  result  found  for  a  side  of  a  regular  twelve-sided 
polygon  substituted  for  A  B  in  the  above  formula,  gives 
the  side  of  a  regular  twenty-four-sided  polygon. 

In  this  way  the  sides  of  regular  inscribed  polygons  of 
greater  and  greater  number  of  sides  may  be  found. 
Some  of  the  results  are  given  in  the  following  table: 


NO.  SIDES. 

ONE   SIDE. 

PERIMETER. 

6 
1$ 

24 
48 
96 
192 

1/2 

-  >/3 

1 
'=.51763809 

=  .26105238 
=  .13080626 
=  .06533817 
-  .03271346 

6 

6.21165708 
6.26525722 
6.27870041 
6.28206396 
6.28290510 

1/2 

-N/4 

-(.51763809)  2 

V2 

-  x/4 

-  (.26105238)  2 

1/2 

-v/4 

-(.13080626)2 

1/2 

-v/4 

-  (.06533817)  2 

REGULAR   POLYGONS  —  CIRCLES.  233 

In  this  way,  the  perimeter  of  a  polygon  of  768  sides 
has  been  computed  to  be  6. 283169 +  . 

Dividing  this  result  by  the  diameter,  i.  e.,  by  2,  gives 
3. 141584 +  ,  as  an  approximate  value  of  the  ratio  of  the 
circumference  of  a  circle  to  its  diameter,  i.  e. ,  an  approx- 
imate value  of  n. 

The  approximate  value  usually  used  is  3.1416. 

Therefore,  n  =  3.1416  approximately. 

323.  COROLLARY.     As  an  approximate  value  of  n  has 
been  found,  the  area  of  a  circle  may  be  found  approx- 
imately in  terms  of  its  radius.     The  approximate  value 
is  found  by  multiplying  the  square  of  the  radius  by  3*14.16. 
(Art.  315.) 

324.  SCHOLIUM.     Archimedes  (born  287  B.  C),  found 
an  approximate  value  of  n.     He  proved  that  its  value  is 
between  3|  and  3fJ.     The  smaller  of  these  two  values  is 
often  used  as  an  approximate  value  of  n  when  great  ac- 
curacy is  not  required. 

In  modern  times  the  value  of  n  has  been  computed  to 
a  large  number  of  decimal  places.  Clausen  and  Dase,  in- 
dependently of  each  other,  computed  the  value  to  the 
two  hundredth  decimal  place.  Other  computers  have 
given  the  value  to  over  five  hundred  decimal  places,  but 
their  results  have  not  been  verified.  The  number  is  in- 
commensurable, and  cannot  be  expressed  exactly  by 
any  number  of  decimal  places. 


Ex.  234.  The  altitude  of  an  equilateral  triangle  equals 
one  and  one  half  times  the  radius  of  the  circumscribed 
circle. 

Ex.  235.  If  the  center  of  each  of  two  equal  circles  lies 
on  the  circumference  of  the  other,  the  square  on  the 
common  chord  is  three  times  the  square  on  the  radius. 


234 


PLANE    GEOMETRY. 


PROPOSITION  XVII.     PROBLEM. 
325.  To  inscribe  a  square  in  a  given  circle. 


Let  Obea  given  circle. 

To  inscribe  a  square  in  the  circle  O.    ' 

SuG.  In  any  square,  at  what  /_  do  the  diagonals  in- 
tersect ?    How,  then,  can  a  square  be  inscribed  in  a  circle  ? 

326.  COROLLARY.     By  bisecting  the  arcs  subtended 
by  the  sides  of  the  square,  and  joining  each  point  of  di- 
vision with  the  two  adjacent  vertices,  a  regular  inscribed 
octagon  is  formed. 

PROPOSITION  XVIII.     PROBLEM. 

327.  To  inscribe  a  regular  hexagon  in  a  given  circle. 


O 


Let  O  be  a  given  circle. 

To  inscribe  a  regular  hexagon  in  the  circle  O. 


REGULAR    POLYGONS  —  CIRCLES.  235 

SUG.  Lay  off  six  times  as  a  chord,  the  line  which  equals 
one  side  of  the  hexagon.     How  can  this  line  be  obtained  ? 

328.  COROLLARY  I.     By  joining  the  alternate  vertices 
of  the  regular  inscribed  hexagon,  an  equilateral  triangle 
is  inscribed  in  the  circle. 

329.  COROLLARY  II.     By  bisecting  the  arcs  subtended 
by  the  sides  of  a  regular  inscribed  hexagon,  and  joining 
the  points  of  division  with  the  adjacent  vertices  of  the 
hexagon,  a  regular  dodecagon  (i.  e.,  a  regular  polygon  of 
twelve  sides),  is  inscribed  in  the  circle. 


Ex/236.  Divide  a  circle  into  segments  such  that  an  angle 
inscribed  in  one  segment  shall  be  three  times  an  angle 
inscribed  in  the  other  segment. 

Ex.  237.  From  a  point  without  a  circle  two  tangents  are 
drawn  which,  with  the  chord  of  contact,  from  an  equi- 
lateral triangle  whose  side  is  18  in.  Find  the  diameter 
of  the  circle. 

Ex.  238.  Find  the  locus  of  the  center  of  a  circle  which 
passes  through  two  fixed  points. 

Ex.  239.  Find  the  locus  of  the  point  of  intersection  of 
tangents  to  a  fixed  circle  which  intersect  at  a  given  angle. 

Ex.  240.  Find  the  locus  of  a  point  equally  distant  from 
the  circumferences  of  two  eqaal  non-intersecting  circles. 

Ex.  241.  Find  the  locus  of  the  center  of  a  circle  tan- 
gent to  two  intersecting  straight  line. 

Ex.  242.  Find  the  locus  of  a  point  equally  distant  from 
the  circumferences  of  two  concentric  circles. 

Ex.  243.  Prove  that  the  apothem  of  an  inscribed  equi- 
lateral triangle  equals  one  half  of  the  radius  of  the  circle. 


236  PLANE    GEOMETRY. 

PROPOSITION  XIX.     PROBLEM. 
330.  To  inscribe  a  regular  decagon  in  a  given  circle. 


Let  Obea  given  circle. 

To  inscribe  a  regular  decagon  in  the  circle  O. 

SUG.  1.  Divide  the  radius  O  A  into  mean  and  extreme 
ratio,  and  let  O  C  be  the  greater  segment. 

SUG.  2.  Draw  the  chord  A  B  equal  to  to  O  C,  and  con- 
nect O  with  B,  and  C  with  B. 

SUG.  3.  See  the  definition  of  mean  and  extreme  ratio 
and  form  a  proportion  with  the  lines  O  A,  O  C  and  C  A. 

SUG.  4.  In  the  proportion  just  formed  substitute  for  O  C 
its  equal,  A  B,  thus  forming  a  new  proportion. 

SUG.  5.  By  means  of  the  last  proportion,  the  As  O  A  B 
and  B  A  C  in  respect  to  similarity.  In  this  comparison 
notice  that  the  Z.  A  is  common  to  the  two  As. 

SUG.  6.  Compare  the  lines  A  B  and  B  C ;  also  the  lines 
O  C  andC£. 

SUG.  7.  Compare  /.ABC  with  Z  O ;  also,  Z  CB  O 
with  Z.  O.  Compare  Z.  A  B  O  with  Z.  O.  Compare 
Z.  B  A  O  with  Z  O. 

SUG.  8.  The  sum  of  the  three  Zs  of  the  A  OB  A  con- 
tains the  Z  O  how  many  times  ? 


REGULAR    POLYGONS  —  CIRCLES.  237 

SUG.  9.  The  Z  O  is  what  part  of  two  rt.  Zs  ?  What 
part  of  four  rt.  Z!s  ?  Hence,  the  chord,  A  B,  subtends 
what  part  of  the  circumference  ? 

SUG.  10.  Now,  give  the  method  of  inscribing  a  regular 
decagon  in  a  circle. 

331.  COROLLARY.     By  joining  the  alternate  vertices 
of  a  regular  inscribed  decagonra  regular  inscribed  penta- 
gon is  formed. 

332.  GENERAL  SCHOLIUM.    Methods  have  already  been 
explained  of  inscribing  in  a  circle  a  regular  polygon  of  3, 
4,  5,  6,  8  or  10  sides.     Any  regular  inscribed  polygon 
being  given  a  regular  inscribed  polygon  of  double  the 
number  of  sides,  can  be  formed  by  bisecting  the  arcs  sub- 
tended by  the  sides  and  joining  the  points  of  division  to 
the  adjacent  vertices  of  the  given  polygon.     Hence,  by 
means  of  the  inscribrd  square  regular  polygons  can  be 
inscribed  of  8,  16,  32,  etc.,  sides;  by  means  of  the  regular 
inscribed  hexagon  regular  polygons  can  be  inscribed  of 
12,  24,  48,  etc.,  sides;  by  means  of  the  regular  inscribed 
decagon  regular  polygons  can  be  inscribed  of  20,  40,  80, 
etc. ,  sides.     There  is  still  one  more  set  of  polygons  which 
can  be  inscribed  in  a  circle.     For,  if  from  any  point  on  the 
circumference  a  chord  be  drawn  eqnal  to  one  side  of  a 
regular  inscribed  hexagon,  and  from  the  same  point  an- 
other chord  be  drawn  equal  to  the  side  of  a  regular  in- 
scribed decagon,  then  the  first  chord  subtends  an  arc 
which  is  one  sixth  of  the  circumference,  and  the  second 
chord  an  arc  which  is  one  tenth  of  a  circumference,  and 
the  difference  between  these  two  arcs  is  one  fifteenth  of 
the  circumference.     Hence,  the   chord  joining  the   ex- 
tremities of  the  two  chords  previously  drawn  is  one  side 
of  a  regular  inscribed  polygon  of  fifteen  sides,  and  from 
this  regular  polygons  can  be  inscribed  of  30,  60,  etc.,  sides. 


238  PLANE   GEOMETRY. 


Until  the  beginning  of  the  present  century  it  was  sup- 
posed that  the  polygons  already  enumerated  were  the 
only  ones  which  could  be  inscribed  by  elementary  geome- 
try, but,  in  a  work  published  in  1801,  Gauss  proved  by 
means  of  the  ruler  and  dividers  only,  it  is  possible  to  in- 
scribe regular  polygons  of  17  sides,  of  257  sides,  and,  in 
general,  of  any  number  of  sides  which  can  be  expressed 
by  2*-f  1,  n  being  an  integer  provided  that  2"-f  1  is  a 
prime  number. 


PROPOSITIONS  IN   CHAPTER  V. 


PROPOSITION  I. 
An  equilateral  polygon  inscribed  in  a  circle  is  a  regular  polygon. 

PROPOSITION  II. 

A  circle  can  be  circumscribed  about  a  regular  polygon,  and  a  circle 
can  be  inscribed  in  a  regular  polygon. 

PROPOSITION  III. 

If  a  circumference  is  divided  into  any  number  of  equal  parts,  the 
tangents  drawn  through  the  points  of  division  form  a  regular  circum- 
scribed polygon. 

PROPOSITION  IV. 
Regular  polygons  of  the  same  number  of  sides  are  similar. 

PROPOSITION  V. 

The  perimeters  of  two  regular  polygons  of  the  same  cumber  of 
sides  have  the  same  ratio  as  their  radii,  or  their  apothems. 

PROPOSITION  VI. 

If  the  number  of  sides  of  a  regular  inscribed  polygon  be  increased 
indefinitely  the  apothem  is  a  variable  which  approaches  the  radius  as 
a  limit. 

PROPOSITION  VII. 

Ii  the  number  of  sides  of  a  regular  inscribed  polygon  be  increased 
indefinitely  the  perimeter  of  the  polygon  is  a  variable  which  ap- 
proaches the  circumference  of  the  circle  as  a  limit. 

PROPOSITION  VIII. 

If  the  number  of  sides  of  a  regular  inscribed  polygon  be  increased 
indefinitely,  the  area  of  the  polygon  is  a  variable  which  approaches 
the  area  of  the  circle  as  a  limit. 


240  PLANE   GEOMETRY. 

PROPOSITION  IX. 

The  area  of  a  regular  polygon  is  equal  to  one  half  the  product  of  its 
perimeter  and  apothem. 

PROPOSITION  X. 

The  circumferences  of  two  circles  have  the  same  ratio  as  their 
radii. 

PROPOSITION  XI. 

The  areas  of  two  circles  have  the  same  ratio  as  the  squares  of  their 
radii. 

PROPOSITION  XII. 

The  area  of  a  circle  equals  one  half  the  product  of  its  circumfer- 
ence and  radius. 

PROPOSITION  XIII. 

One  side  of  a  regular  hexagon  equals  the  radius  of  the  circum- 
scribed circle. 

PROPOSITION  XIV. 

The  areas  of  two  similar  segments  have  the  same  ratio  as  the 
squares  of  their  radii. 

PROPOSITION  XV. 

Given  the  radius  of  a  circle  and  the  side  of  a  regular  inscribed 
polygon,  required  to  find  the  side  of  a  regular  inscribed  polygon  of 
double  the  number  of  sides. 

PROPOSITION  XVI. 

To  compute,  approximately,  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter. 

PROPOSITION  XVII. 
To  inscribe  a  square  in  a  given  circle. 

PROPOSITION  XVIIL 
To  inscribe  a  regular  hexagon  in  a  given  circle 

PROPOSITION  XIX. 
To  inscribe  a  regular  decagon  in  a  given  circle. 


GEOMETRY  OF  THREE  DIMENSIONS. 


CHAPTER  VI. 
LINES  AND  PLANES. 


DEFINITIONS. 

333.  Geometry  of  three  dimensions  treats  of  fig- 
ures whose  parts  are  not  confined  to  a  single  plane. 

Geometry  of  three  dimensions  is  also  called  solid 
geometry  and  geometry  of  space. 

The  results  of  plane  geometry  furnish  the  basis  for  the 
investigations  of  geometry  of  three  dimensions,  but  it 
must  be  remembered  that  the  statements  of  plane  geome- 
try are  made  with  respect  to  figures  which  are  entirely  in 
one  plane,  and,  in  the  form  stated,  are  not  necessarily  true 
in  geometry  of  three  dimensions.  For  example,  in  plane 
geometry  it  is  stated  that  from  a  given  point  in  a 
straight  line  only  one  perpendicular  can  be  drawn  to 
that  line.  This  statement  is  true  only  of  figures  confined 
to  a  single  plane,  and  hence  is  not  true  in  geometry  of 
three  dimensions. 

In  reasoning  from  plane  geometry  to  geometry  of  three 
dimensions,  the  following  axiom  is  used. 

334.  Axiom  14.  The  relations  of  the  parts  of  a  figure, 
or  figures,  in  one  plane  are  not  changed  by  moving  the 
plane   containing  the  figure  from  one  position  to   an- 
other. 

16— Geo. 


242  SOLID   GEOMETRY. 

335.  A  plane  has  been  defined  as  a  surface  such  that  a 
straight  line  joining  any  two  of  its  points  will  lie  wholly 
in  the  surface. 

A  plane  is  considered  to  be  indefinite  in  extent,  so  that 
if  any  straight  line  of  the  plane  be  extended  indefinitely 
it  will  continue  to  lie  wholly  in  the  surface. 

336.  A  plane  is  said  to  be  determined  when  it  is  ex- 
actly located,  or  is  distinguished  from  every  other  plane. 
A  plane  is  determined  by  lines  or  points  when  it  is  the 
only  plane  which  contains  those  lines  or  points. 


Ex.  244.  The  apothem  of  a  regular  inscribed  hexagon 
equals  ^  R  1/3  ;  R  representing  the  radius  of  the  circle. 

Ex.  245.  One  side  of  an  inscribed  equilateral  triangle 
equals  R  1/3  ;  R  being  the  radius  of  the  circle. 

Ex.  246.  One  side  of  an  inscribed  square  equals  R  1/2  , 
and  its  apothem  equals  \  R  1/2;  R  being  the  radius  of  the 
circle. 

Ex.  247.  Show  how  to  cut  off  the  corners  of  an  equi- 
lateral triangle  in  such  a  way  that  the  remaining  figure 
will  be  a  regular  hexagon. 

Ex.  248.  If  any  two  polygons  whatever  be  circum- 
scribed about  the  same  circle,  prove  that  their  areas  have 
the  same  ratio  as  their  perimeters. 

Ex.  249.  In  a  given  circle  inscribe  an  isosceles  triangle 
in  which  each  base  angle  is  twice  the  vertical  angle. 

Ex.  250.  In  a  given  circle  inscribe  a  triangle  whose 
angles  are  proportional  to  the  numbers  3,  4  and  5. 

Ex.  251.  What  is  the  area  of  a  sector  whose  arc  is  one 
sixth  of  the  circumference,  in  a  circle  whose  radius  is 
15  in. 


LINES    AND   PLANES.  243 

POPOSITION  I.     THEOREM. 

337.  A  plane  may  be  passed  through  any  straight 
line,  but  the  line  does  not  determine  the  plane. 


.D 


Let  A  I?  represent  any  straight  line. 

To  prove  that  a  plane  may  be  passed  through  the  line  A  B, 
but  that  the  line  does  not  determine  the  plane. 

Suo.  1.  Consider  any  plane  whatever,  and  draw  any 
straight  line  in  the  plane. 

SUG.  2.  Move  the  line  and  plane  until  the  line  of  the 
plane  coincides  with  the  given  line,  A  B. 

SUG.  3.  Is  it  possible,  then,  to  pass  a  plane  through 
the  line  A  B1 

SUG.  4.  Must  the  plane  always  occupy  exactly  the 
same  position  when  the  line  of  the  plane  coincides 
with  A  fit 

SUG.  5.  Does  the  line  determine  the  plane? 

Therefore 

338.  When  a  plane  takes  in  succession  the  various 
possible  positions  with  the  line  of  the  plane  coinciding 
with  the  line  A  B,  the  plane  is  said  to  revolve  about  A  B. 


244:  SOLID    GEOMETRY. 

PROPOSITION  II.     THEOREM. 

339.  A  plane  is  determined: 

I.     By  a  straight  line  and  a  point  without  the  line* 
II.     By  two  intersecting  straight  lines. 

III.  By  two  parallel  straight  lines. 

IV.  By  three  points  not  in  a  straight  line. 


C 


* 

l\r 


w 

Let  A  B  represent  a  straight  line,  and  C  a  point  not  in 
the  line  A  B. 

To  prove  that  the  line  A  B  and  the  point  C  determine  a 
plane. 

SUG.  1.  Through  the  line  A  B  pass  a  plane,  and  let  it 
revolve  about  A  B  until  it  contains  the  point  C. 

SUG.  2.   How  much  can  the  plane  be  revolved,  either 
way,  about  B,  and  still  contain  the  point  C  ?     Why  ? 

SUG.  3.  What  is  your  conclusion  about  part  I  of  the 
proposition  ? 

M 


/ 


Let  A  B  and  C  D  represent  two  intersecting  lines. 

To  prove  that  A  B  and  C  D  determine  a  plane. 

SUG.  1.  Pass  a  plane,  M '  N,  through  the  line  A  B  and 
some  point  of  C  D. 

SUG.  2.  How  many  planes  can  occupy  this  position  ? 
Why? 


LINES    AND    PLANES.  245 

SuG.  3.  Where  does  the  line  C  D  lie  with  respect  to 
the  plane  M  Nt  (Art.  335.)  How  many  points  of  C  D 
are  in  the  plane  M  Nt 

SuG.  4.  Do  the  lines  A  B  and  C  D  determine  a  plane  ? 


-.' 


Let  A  B  and  C  D  represent  two  parallel  lines. 

To  prove  that  A  B  and  C  D  determine  a  plane. 

SuG.  1.  By  definition  of  ||  lines,  in  how  many  planes 
do  they  lie  ? 

SuG.  2.  Pass  a  plane  through  A  B,  and  one  point 
of  C  D.  How  many  planes  occupy  this  position  ?  Where 
is  the  plane  of  ||  lines  located  with  respect  to  this  plane  ? 
Then,  how  many  planes  can  pass  through  the  lines  A  B 
andC£>? 


"Let  Af  B   and  C  represent  three  points   not  in  one 

straight  line. 

To  prove  that  the  points  A,  B  and  C  determine  a  plane. 

SUG.  Connect  two   of  the   points.     What   conditions 
have  you  now  ? 
Therefore 


246  SOLID   GEOMETRY. 

PROPOSITION  III.    THEOREM. 

340.  The  intersection  of  two  planes  is  a  straight 
line. 


\/       \/ 
J\ 


Let  MN  and  C  D  represent  two  planes,  and  A  and  B 
two  points  of  their  intersection. 

To  prove  that  M  N  and  CD  intersect  in  a  straight  line. 

SuG.  1.  Connect  A  and  B.  Where  is  the  line  A  B 
with  respect  to  each  of  the  planes  ? 

SuG.  2.  Take  any  point  of  the  plane  M  N,  without  the 
line  A  B.  Can  this  point  be  in  the  plane  C  D  also  ? 
Why  ?  (See  Prop.  II,  part  I.) 

Therefore 

QUERY.  What  is  the  locus  of  a  point  common  to  two 
planes  ? 

341.  The  point  in  which  a  straight  line  intersects  a 
plane  is  called  the  foot  of  the  line. 

342.  A  line  is  perpendicular  to  a  plane  when  it  is  per- 
pendicular to  every  line  in  the  plane  drawn  through  its 
foot. 

The  plane  is  then  said  to  be  perpendicular  to  the  line. 


LINES    AND    PLANES. 


247 


PROPOSITION  IV.     THEOREM. 

343.  Fj'om  a  given  point  without  a  plane,  one,  and 
only  one,  perpendicular  can  be  dropped  to  the  plane, 
and  the  perpendicular  is  the  shortest  line  from  the 
point  to  the  plane. 


Let  M  N  represent  the  given  plane,  and  A  the  given 
point  ivithout  the  plane. 

To  prove  that  y  from  Ay  one,  and  only  one,  perpendicular 
can  be  drawn  to  the  plane  M  JVt  and  that  the  perpendicular 
is  shorter  than  any  other  line  from  A  to  the  plane  M  N. 

SUG.  1.  Of  all  lines  from  A  to  the  plane  M  N,  either 
there  is  one  shortest  line  or  a  group  of  equal  shortest 
lines.  Suppose  A  C  and  A  D  are  two  of  a  group  of 
shortest  lines.  Connect  C  and  D. 

SUG.  2.  If^Cand^Z?are=,whatkindof  A  is  ^  CD? 

SuG.  3.  From  A,  draw  A  B  _L  to  CD.  Compare  A  B 
and  A  C  in  respect  to  length. 

SUG.  4.  If  any  two  lines  from  A  to  the  plane  M  N,  are 
equal  in  length,  can  they  be  shortest  fines  to  the  plane  ? 

SUG.  5.  Then,  how  many  shortest  lines  are  there  from 
A  to  the  plane  M  N? 

SUG.  6.  Let  A  O  represent  the  shortest  line  from  A  to 
the  plane  M  N,  and  draw  E  G,  any  line  in  the  plane  M  N, 
through  the  point  O. 

SUG.  7.  Since  A  O  is  the  shortest  line  from  A  to  the 
line  E  G,  what  relation  does  A  O  sustain  to  E  G  ?  Why  ? 

SuG.  8.  Since  E  G  is  any  line  drawn  through  the  foot 
of  A  O,  what  relation  must  A  O  sustain  to  the  plane  M  N1 

SUG.  9.  How  many  J_s  can  be  drawn  from  A  to  the 
plane  M  N1  Why? 

Therefore 


SOLID    GEOMETRY. 


PROPOSITION  V.     THEOREM. 

344.  At  a  given  point  in  a  plane,  one  perpendicu- 
lar, and  only  one,  can  be  erected  to  the  plane. 


.C 


Let  C  D  represent  a  plane,  and  O  a  given  point  in  the 
plane. 

To  prove  that  one  and  only  one  perpendicular  can  be 
erected  to  the  plane  at  the  point-  O. 

SUG.  1.  L,et  M  N  represent  another  plane,  and  A  B 
a  _L  from  A  to  the  plane  M  N. 

SUG.  2.  Place  the  plane  M  N  on  the  plane  C  D,  so 
that  B  coincides  with  O,  When  so  placed,  what  rela- 
tion does  AB  sustain  to  the  plane  CD?  (Ax.  14.) 
Can  a  _L  be  erected  to  the  plane  C  D,  at  the  point  O  ? 

SuG.  3.  If  more  than  one  J_  can  be  erected,  let  O  E 
and  O  G  represent  two  J_s  to  C  D,  at  O.  I^et  R  S  be  the 
intersection  of  the  plane  BOG  with  the  plane  C  D. 
What  relation  do  both  O  E  and  O  G  sustain  to  RSI 
Why  ?  Can  a  second  J_  be  erected  to  a  plane  at  a  given 
point  of  the  plane  ? 

Therefore 


LINES    AND    PLANES.  249 

PROPOSITION  VI.     THEOREM. 

345.  If,  *from  any  point  in  the  perpendicular  to  a 
planet  oblique  -lines  be  drawn,  those  which  meet  the 
plane  at  equal  distances  from  the  foot  of  the  perpen- 
dicular are  equal ;  and,  of  two  unequal  oblique  lines, 
that  which  meets  the  plane  at  the  greater  distance  from 
the  foot  of  the  perpendicular  is  greater. 


Let  M  N  represent  a  plane,  and  A  B  a  perpendicular  to 
the  plane ;  also,  let  A  D,  A  C  and  A  E  represent  oblique 
lines  meeting  tfie  plane  MN  at  the  points  J>,  C  and  E  re- 
spectively, and  let  B  C  equal  B  E,  and  B  D  be  greater 
tftan  B  E. 

To  prove  that  the  oblique  lines  A  C  and  A  E  are  equal \ 
and  that  the  line  A  D  is  longer  than  the  line  A  E. 

SUG.  1.  What  relation  does  A  B  sustain  to  B  C,  and 
also  to  B  E  ?  Why  ?  (Art.  342.) 

SUG.  2.   Compare  As  A  B  C  and  ABE. 

SUG.  3.   Compare  A  C  and  A  E. 

SUG.  4.  Revolve  the  plane  of  the  A  A  B  E  upon  A  B 
as  an  axis,  into  the  plane  of  A  B  D,  and  let  A  B  F  be 
the  position  which  ABE  takes  after  the  revolution. 

SUG.  5.  Since  Zs  A  B  F  and  A  B  D  are  rt.  Zs  (why 
are  they  rt.  Zs  ?),  where  will  B  F  lie  with  respect  to  B  D  ? 

SUG.  6.  Compare  A  F  with  A  Dt  and  hence  A  E 
with  A  D. 

Therefore 


250 


SOLID    GEOMETRY. 


PROPOSITION  VII.     PROBLEM. 

346.  If  ou  straight  line  is  perpendicular  to  two  lines 
of  a  plane  at  their  point  of  intersection  it  is  perpen- 
dicular to  the  plane. 


Let  My  represent  a  plane,  A  U  and  C  J>  two  lines  of 
the  plane,  and  E  O  a  perpendicular  to  the  two  lines  at 
their  point  of  intersection,  O. 

To  prove  that  O  E  is  perpendicular  to  the  plane  M  N. 

SuG.  1.  Extend  O  E  to  S,  making  O  S  =  E  O,  and 
let  O  R  represent  any  line  of  the  plane  through  the  point 
O.  Draw  the  line  B  Z>,  intersecting  O  R  at  R.  Con- 
nect both  E  and  S  with  the  points  B,  R  and  D. 

SuG.  2.  In  As  E  O  B  and  SOB,  compare  E  B  and 
SB\  also,  m&sEOD  and  SOD  compare  ED  and  SD. 

SuG.  3.  In  As  E  B  D  and  S  B  D,  compare  Z.  E  B  D 
with  Z.SB  D. 

SuG.  4.   In  As^^^andS^^compare.E^with.S^. 

SUG.  5.  In  As  E  O  R  and  S  O  R,  compare  Z_$  E  O  R 
and  SO  R. 

SuG.  6.  What  relation  does  E  O  sustain  to  O  R  ? 

SUG.  7.  Since  O  R  is  any  line  of  the  plane  through 
the  point  O,  what  relation  does  E  O  sustain  to  the  plane  ? 
(Art.  342.) 

Therefore 


LINES    AND    PLANES.  251 

PROPOSITION  VIII.     THEOREM. 

347.  All  the  perpendiculars  to  a  given  line  at  the 
same  point  lie  in  the  same  planet  and  that  plane  is 
perpendicular  to  the  given  line. 


B 


Let  BO  be  a  given  line,  and  O  A,  O  C,  O  J>,  etc.,  lines 
peri>emlicular  to  &  O,  at  O. 

To  prove  that  OA,  O  C,  O  D,  etc.,  are  in  the  same  plane, 
and  that  the  plane  is  perpendicular  to  B  O,  at  O. 

SUG.  1.  The  lines  O  D  and  O  C  determine  a  plane. 
The  lines  O  B  and  O  A  determine  another  plane.  Let 
these  two  planes  intersect  in  the  line  O  M. 

SuG.  2.  What  relation  does  B  O  sustain  to  the  plane 
determined  by  O  D  and  O  C?  Why  ?  (Art.  346.) 

SUG.  3.  What   relation   does    OB   sustain   to 
Why? 

SUG.  4.  What  relation  does  O  B  sustain  to  O  A  ? 

SuG.  5.  What  relation  then  must  O  A  sustain  to 

SUG.  6.  Then,  where  must  O  A  lie  with  respect  to  the 
plane  determined  by  O- D  and  O  C? 

Therefore 


252  SOLID    GEOMETRY. 

348.  COROLLARY  I.     Through   a   given   point   in    a 
straight  line  only  one  plane  can  be  drawn  perpendicular 
to  the  line. 

Suppose  one  plane  drawn  through  O,  _L  to  O  B. 
Then,  if  a  second  plane  can  be  drawn  through  O  J_  to 
O  B,  all  the  lines  in  that  second  plane  through  O  would 
be  _L  to  O  B,  at  O.  But  all  the  lines  _L  to  OB  at  O  are 
in  the  same  plane.  Hence,  the  second  plane  must  coin- 
cide with  the  first.  That  is,  only  one  plane  can  be  drawn 
through  O  _L  to  B  O. 

349.  COROLLARY  II.     From  a  given  point  without  a 
line  only  one  plane  can  be  drawn  perpendicular  to  the 
given  line. 


X~  -*B 

the  plane  A  B,  through  the  point  Ny  be  _L  to  M  O 
at  O. 

If  another  plane  can  be  drawn  through  N  J_  to  M  O, 
represent  it  by  C  B,  and  suppose  C  B  is  _L  to  .M  O  at  5. 
Connect  N  and  O ;  also  N  and  S. 

What  relation  must  ON  and  5  N  each  sustain  to  MS? 
Why? 


Ex.  252.  Find  the  area  of  a  circular  ring  included  be- 
tween the  circumferences  of  two  concentric  circles,  whose 
radii  are  3  and  6  respectively. 


LINES    AND    PLANES.  253 

PROPOSITION  IX.     THEOREM. 

350.  If,  from  the  foot  of  a  perpendicular  to  a  plane, 
a  line  be  drawn  perpendicular  to  any  line  of  tJie 
plane,  and,  from  the  point  of  intersection,  a  line  be 
drawn  to  any  point  of  the  perpendicular,  the  last  line 
is  perpendicular  to  the  line  of  the  plane. 


S^}D 

Ssr  / 

--4L 


Let  M  N  represent  a  plane,  and  A  B  a  perpendicular  to 
tJie  plane.  Let  C  D  represent  a  line  of  the  plane  MN,  and 
let  B  O  be  perpendicular  to  I)  C,  at  O,  and  let  the  point  of 
intersection  O,  be  joined  to  F,  a  point  in  A  B. 

To  prove  that  F  O  is  perpendicular  to  C  D. 

SUG.  1.  On  the  line  D  C,  take  O  D  =  O  C     Connect 
F  with  C  and  D ;  also  connect  B  with  C  and  D. 
SUG.  2.  Compare  As  B  O  D  and  £  O  C. 
SUG.  3.  Compare  B  D  with  B  C. 
SUG.  4.  Compare  As  FB  D  and  F  B  C. 
SUG.  5.  Compare  F  C  and  F  D. 
SUG.  6.  Compare  As  F  C  O  and  F  D  O. 
SUG.  7.  Compare  Zs  D  O  /^and  C  O  F. 
Therefore 


254 


SOLID    GEOMETRY. 


351.  A  straight  line  is  parallel  to  a  plane  when  the 
line  and  plane  cannot  meet  however  far  they  may  both 
be  extended. 

In  that  case,  the  plane  is  also  parallel  to  the  line. 

352.  Two  planes  are  parallel  to  each  other  when  they 
cannot  meet  however  far  both  may  be  extended. 

PROPOSITION  X.    THEOREM. 

353.  If  one  of  two  parallel  straight  lines  is  perpen- 
dicular to  a  plane,  the  other  is  perpendicular  to 
the  same  plane. 


Let  A  J5  and  CD  be  two  parallel  lines,  meeting  the  plane 
MN in  the  points  B  and  D  respectively,  and  let  CD  be 
perpendicular  to  the  plane  MN. 

To  prove  that  A  B  is  perpendicular  to  the  plane  M  N. 

SUG.  1.  Connect  B  and  Z>,  and,  through  B,  draw  a  line 
E  Ft  in  the  plane  M  N,  JL  to  B  D.  Connect  B  with  any 
point  O  in  CD. 

SUG.  2.  What  relation  does  OB  sustain  to  EF,  or  EF 
to  OBt  What  relation  does  EF  sustain  to  B  Dt 

SUG.  3.  What  relation  does  E  F  sustain  to  the  plane 
ABDC1  (Art.  346.)  What  relation  does  E  F  sustain 
to  A  Bt 


LINES    AND    PLANES. 


255 


SUG.  4.  What  relation  does  A  B  sustain  to  E F*  What 
relation  does  A  B  sustain  to  B  Z>?  Why  ?  What  rela- 
tion does  A  B  sustain  to  the  plane  M  N1  Why  ? 

Therefore 

PROPOSITION  XI.    THEOREM. 

354.  If  two  lines  are  perpendicular  to  the  same 
plane,  they  are  parallel. 


Li€t  the  two  straight  lines  A  B  and  CD  be  perpendicular 
to  tJie  plane  M  N. 

To  prove  that  A  B  and  C  D  are  parallel. 

SUG.  1.  If  A  B  is  not  ||  to  CD,  draw,  through  any 
point  <9,  of  A  B,  a  line  O  E  ||  to  C  D. 

SUG.  2.  What  relation  does  O  E  sustain  to  the  plane 
Mm  (Art.  253.) 

SuG.  3.  Compare  the  answer  to  Sug.  2  with  the  hy- 
pothesis, and  finish  the  demonstration. 

Therefore 


Ex.  253.  Find  the  line  on  which  a  paper  triangle  must 
be  folded  in  order  that  the  vertex  may  fall  upon  a  given 
point  of  the  base. 


256 


SOLID    GEOMETRY. 


PROPOSITION  XII.    THEOREM. 

355.  If  two  straight  lines  are  each  parallel  to  a 
third  straight  line  they  are  parallel  to  each  other. 


D 


B 


Let  A  B  and  C  J>  each  be  parallel  to  E  F. 

To  prove  that  A  B  and  C  D  are  parallel  to  each  other. 

SUG.  1.  Draw  a  plane,  M  N,  _L  to  E  P. 

SUG.  2.  What  relation  does  A  B  sustain  to  the  plane 
M N  ?  What  relation  does  CD  sustain  to  the  plane  MNt 

SUG.  3.  What  relation  do  A  B  and  C  D  sustain  to 
each  other  ? 

Therefore 


Ex.  254.  Equal  oblique  lines,  drawn  from  a  point  in 
the  perpendicular  to  a  plane,  meet  the  plane  at  equal  dis- 
tances from  the  foot  of  the  perpendicular,  and  of  two  un- 
equal oblique  lines  the  longer  meets  the  plane  at  the 
greater  distance  from  the  foot  of  the  perpendicular. 

Ex.  255.  What  is  the  locus  of  the  foot  of  an  oblique 
line  of  constant  length  drawn  from  a  point  in  a  perpen- 
dicular to  a  plane  ? 


LINES    AND    PLANES. 


257 


PROPOSITION  XIII.     THEOREM. 

356.  Every  plane  containing  one,  and  only  one,  of 
two  parallel  lines,  is  parallel  to  the  other  line. 
C 


tJ, 


Let  A  B  and  CD  be  two  parallel  lines,  and  MNa  plane 

<-<>ni(ihtin<f  thr  fine  A  B. 

To  prove  that  the  line  C  D  and  the  plane  M  N  are  par- 
allel. 

SUG.  1.  A  B  and  C  D  determine  a  plane  C  B.     Why  ? 

SUG.  2.  What  is  the  intersection  of  the  planes  M  N 
andC^?  Why? 

SUG.  3.  If  CD  meets  the  plane  M  N,  it  must  meet  it 
in  the  line  A  B.  Why  ? 

SUG.  4.  Is  it  possible  for  BD  to  meet  the  plane  M '  Nt 
Why? 

Therefore 


-B 


357.  COROLLARY  I.  Through 
a  given  straight  line  a  plane  can 
be  passed  parallel  to  any  given 
straight  line. 

Through  CD  a  plane  can  be 
passed  parallel  to  A  B.  ~~~N 

SUG.  From  any  point  in  CD,  draw  a  line  ||  to  A  B ;  as 
the  line  O  P.     The  plane  determined  by  C  D  and  O  P  is 
parallel  to  A  B.     Why  ? 
17— Geo. 


258  SOLID    GEOMETRY. 

358.  COROLLARY  II.     A  plane  can  be  passed  parallel 
to    any    two   straight    lines 

through  any  given  point  with- 
out the  lines. 

Let  A  B  and  C  D  be  any 
two  lines,  and  O  any  point 
without  the  lines. 

SUG.  Through  6>,  draw  two 
lines  ||  to  A  B  and  CD  re- 
spectively. 

PROPOSITION  XIV.     THEOREM. 

359.  If  a  line  and  a  plane  are  parallel,  the  inter- 
section of  the  plane  with  any  plane  containing  the 
line  is  parallel  to  the  line. 


{ 
£- 


1, 


Let  the  line  CD  be  parallel  to  the  plane  MN,  and  let  the 
plane  C  B,  containing  the  line  C  D,  intersect  M  N  in  the 
line  A  B. 

To  prove  that  A  B  and  C  D  are  parallel. 

QUERY.  Of  what  proposition  is  this  the  converse  ? 

SUG.  Give  two  demonstrations,  a  direct  and  an  iiidi- 
direct. 

Therefore 


LINES    AND   PLANES.  259 

PROPOSITION  XV.     THEOREM. 

360.  Planes  perpendicular  to  the  same  straight  line 
cure  parallel. 

J* 

M 


>j       -IT        7 

o 

/ 

Let  the  planes  MN  and  O  P  be  perpendicular  to  the 


To  prove  that  the  planes  M  N  and  O  P  are  parallel. 

SUG.  1.  Let  C  and  D  be  the  points  of  intersection  of 
the  line  A  B  with  the  planes  M  N  and  O  P  respectively. 

SUG.  2.  If  the  planes  M  N  and  O  P  intersect,  connect 
any  point  of  their  intersection  with  the  points  C  and  D. 

SUG.  3.  What  relation  do  these  lines  sustain  to  A  B  ? 
Why? 

SUG.  4.  Is  this  possible  ? 

Therefore  - 


Ex.  256.  If,  from  a  point  in  a  perpendicular  to  a  plane, 
a  line  be  drawn  perpendicular  to  any  line  of  the  plane,  a 
line  joining  the  point  of  intersection  to  the  foot  of  the 
perpendicular  to  the  plane  is  perpendicular  to  the  line  of 
the  plane. 

SUG.  Use  the  indirect  method. 

QUERY.  Of  what  proposition  is  this  exercise  the  con- 
verse ? 


260 


SOLID    GEOMETRY. 


PROPOSITION  XVI.     THEOREM. 

361.  The  intersections  of  two  parallel  planes  with  a 
third  plane  are  parallel  lines. 


Let  MN  and  O  P  represent  two  parallel  planes,  and  A  B 
a  plane  intersecting  them  in  the  lines  C  Z>  and  E  F. 

To  prove  that  C  D  and  E  F  are  parallel  lines, 
SUG.   C  D  and  E  F  are  in  same  plane  A  B.     Can  they 
meet  ?     Why  ? 
Therefore  - 

PROPOSITION  XVII.     THEOREM. 

362.  If  two  angles  not  in  the  same  plane  have  their 
sides  parallel  and  lying  in  the  same  direction  from 
their  vertices,  they  are  equal  and  their  planes  are  par- 
allel. 

AL 


Let  B  AC  and  E  D  F  represent  two  equal  angles  in 
the  planes  M  N  and  O  F  respectively,  having  their  sides 
parallel  and  lying  in  the  same  direction  from  their  vertices. 


LINES    AND    PLANES.  261 

To  prove  that  the  angles  BAG  and  E  D  F  are  equal, 
and  that  the  planes  M  N  and  O  P  are  parallel. 

SUG.  1 .  Take  D  E  =  A  B,  and  D  F  =  A  C.  Con- 
nect A  and  D,  C  and  F,  B  and  E,  C  and  B,  F  and  E. 

SUG.  2.  What  kind  of  a  quadrilateral  is  A  C  F  Dl 
Why  ?  What  kind  of  a  quadrilateral  is  A  B  E  D  ? 

SUG.  3.  Compare  ^  Z?  with  C  F,  also  ^  Z>  with  B  E. 
Give  auth.  Compare  C  F  with  ^  E.  (Art.  355,  and 
Ax.  1.) 

SUG.  4.   Compare  C/?  with  F  E.     Give  auth. 

Sue.  5.  Compare  As  B  A  C  and  E  D  F.  Compare 
Z  B  A  C  with  Z.E  D  F. 

SUG.  6.  What  relation  does  the  plane  (9  />  sustain  to 
the  line  A  B  ?  Also  to  the  line  A  C  ?  Also  to  the  line 
B  C?  Why?  (Art.  356.) 

SUG.  7.  If  the  planes  M  N  and  OP  intersect,  their 
line  of  intersection  is  ||  to  A  B.  Why  ?  (See  Prop.  XIV.) 
Also,  their  line  of  intersection  is  ||  to  A  C.  Why  ? 

Is  it  possible  for  the  planes  M  N  and  O  P  to  intersect  ? 

Therefore 


Ex.  257.  Find  the  locus  of  a  point  equidistant  from 
two  given  parallel  planes. 

Ex.  258.  What  is  the  locus  of  a  point 'in  space  equi- 
distant from  two  given  points  ? 

Ex.  259.  What  is  the  locus  of  a  point  equidistant  from 
two  given  parallel  planes  and  at  the  same  time  equidis- 
tant from  two  given  points  ? 

Ex.  260.  What  is  the  locus  of  a  point  equidistant  from 
two  given  points,  and  at  the  same  time  equidistant  from 
two  other  given  points  ? 


262  SOLID    GEOMETRY. 

PROPOSITION  XVIII.    THEOREM. 

363.  A  straight  line  perpendicular  to  one  of  two 
parallel  planes  is  perpendicular  to  the  other  also. 


•  f  s^  C-  --#      / 

M 


'  A'  77 

^I^^'^^C^ 
0'~ 


Let  MNand  O  P  represent  two  parallel  planes,  and  let 
the  line  ABbe  perpendicular  to  the  plane  MN.  > 

To  prove  that  the  line  A  B  is  perpendicular  to  the 
plane  O  P. 

SuG.  1.  Through  A,  in  the  plane  MN,  draw  two  lines 
CD  and  G  H ' ;  and,  through  B,  in  the  plane  O  P,  draw 
the  lines  E  F  and  K  L  ||  to  C  D  and  G  H  respectively. 
Is  this  construction  possible  ?  Why  ? 

SuG.  2.  What  relation  does  A  B  sustain  to  E  F? 
What  relation  does  A  B  sustain  to  K  LI  Give  auth. 

SUG.  3.  What  relation  does  A  B  sustain  to  the  plane 
OP?  Give  auth. 

Therefore 


Ex.  261.  What  is  the  locus  of  a  point  equidistant  from 
two  given  parallel  planes,  and  at  the  same  time  equidis- 
tant from  two  other  parallel  planes  ? 

Ex.  262.  What  is  the  locus  of  a  point  equally  distant 
from  three  given  points  which  are  not  in  a  straight  line  ? 


LINES    AND    PLANES.  263 

PROPOSITION  XIX.     THEOREM. 

364.  If  three  parallel  planes  are  intersected  by  two 
lines  the  segments  of  the  lines  are  proportional. 

v     fc 

K A.,     ^r 


^A        / 

~V~     \      ~ls 

**        iir        ~^\~ 

Let,  M  N,  OP  and  R  S  represent  three  parallel  planes, 
and  A  B  and  C  D  two  lines  intersectiny  them,  making 
senments  E  I  and  I L  on  A  Bf  and  K  H  and  H  F  on  CD. 

To  prove  that   -=-=-  =  . 

/  L,         ri  r 

SUG.  1.  Connect  E  and  F,  and  let  the  line  E  F  inter- 
sect the  plane  O  Pat  G. 

SuG.  2.  The  plane  E  F  K  intersects  the  planes  M  N 
and  O  P  in  what  lines  ? 

SuG.  3.  The  plane  E  F  L  intersects  the  planes  OP 
and  R  S  in  what  lines  ? 

SUG.  4.  In  the  A  E  F  K,  the  lines  G  H  and  E  K  are 
how  related?  (Prop.  XVI.) 

F  (^  PC  f-f 

SUG.  5.   Compare  the  ratios  and  - 

(JT  r1  Mr 

SUG.  6.   In    a    similar    manner    compare    the    ratios 

EG        ,  El 
-=— =c  and 


GF  IL 

SUG.  7.  C 
Therefore 


F  T  ff  f-f 

SUG.  7.   Compare  the  ratios  -^-f-  and  -Y 

1  L, 


264 


SOLID    GEOMETRY. 


DIHEDRAL  ANGLES. 

365.  Two  planes  which  meet  and  are  terminated  by 
their  line  of  intersection  are  said  to 

form  a  dihedral  angle.  The  planes 
A  B  and  C  D,  in  the  figure  at  the 
right,  meet  in  the  line  A  C  and  form 
a  dihedral  angle. 

The  planes  A  B  and  C  D  are  called 
the  faces  of  the  angle,  and  the  line  A  C 
is  called  the  edge  of  the  angle. 

A  dihedral  angle  is  read  by  reading 
first,  one  face;  second,  the  edge;  third,  the  other  face. 
This  can  be  done  by  reading  four  letters,  if  none  are  re- 
peated; as  B  A  CD.  When  but  one  dihedral  angle  is 
formed  at  an  edge  it  may  be  read  by  reading  the  edge,  as 
dihedral  angle  A  C  in  the  above  figure. 

366.  The  plane   angle  of  a  dihedral  angle  is  an 
angle  formed  by  drawing  two  lines,  one  in  each  face, 
perpendicular  to  the  edge  at  the  same  point. 

In  the  above  figure,  if  F  E  and  F  G  are  perpendicular 
to  the  edge  A  C,  the  angle  E  F  G  is  the  plane  angle  of 
the  dihedral  angle. 

367.  A  plane  is  perpendicular  to  another  plane  if  it 
forms  with  the  other  plane  a  right  dihedral  angle,  z.  <?.,  a 
dihedral  angle  whose  plane  angle  is  a  right  angle. 

368.  Dihedral  angles  are  adjacent,  vertical,  acute,  ob- 
tuse, etc.,   under  the  same  conditions  in  which  plane 
angles  are  adjacent,  vertical,  obtuse,  e.tc.,  respectively. 


LINES    AND    PLANES.  265 

PROPOSITION  XX.     THEOREM. 

369.  All  plane  angles  of  the  same  dihedral  angle 
are  equal. 


Let  ABC  and  D  E  F  be  two  plane  angles  of  the  same 
dihedral  angle. 

To  prove  that  the  angles  ABC  and  DBF  are  equal. 

SUG.  See  Prop.  XVII. 
Therefore 


Ex.  263.  Two  lines  are  perpendicular  to  the  same 
plane.  A  is  a  point  in  one  line  and  B  a  point  in  the  other 
line  such  that  A  and  B  are  equally  distant  from  the  plane. 
Prove  that  the  line  A  B  is  parallel  to  the  plane. 

Ex.  264.  M  N  and  O  P  are  two  parallel  planes.  A  is 
a  point  in  M  N,  and  B  a  point  in  O  P,  such  that  the  line 
A  B  is  perpendicular  to  the  plane  M  N.  C  is  another 
point  in  M ' N,  and  D  another  point  in  O  P,  such  that  the 
line  C  D  is  perpendicular  to  the  plane  O  P.  Prove  that 
the  line  joining  A  with  C  is  parallel  to  the  line  joining 
B  with  D. 


266 


SOLID    GEOMETRY. 


PROPOSITION  XXI.     THEOREM. 

370.  If  a  plane  be  passed  perpendicular  to  the  edge 
of  a  dihedral  angle,  the  lines  of  intersection  with  the 
faces  form  the  plane  angle  of  the  dihedral  angle. 


M 


Let  MNbea  plane  perpendicular  to  the  edge  ABof  a 
dihedral  angle,  and  let  the  lines  of  intersection  form  the 
COD. 


To  prove  that  COD  is  the  plane  angle  of  the  dihedral 
angle. 

SUG.  If  the  plane  M  Ni$  JL  to  the  line  A  B,  what  re- 
lation must  O  C  and  O  D,  in  the  plane  M  N,  sustain 
to  A  El 

Therefore 

371.  Two  dihedral  angles  are  equal  when  they  can  be 
made  to  coincide. 

The  magnitude  of  a  dihedral  angle  does  not  depend 
upon  the  extent  of  its  faces. 

If  a  plane  be  made  to  revolve,  from  the  position  of  one 
face,  about  the  edge  as  an  axis,  to  the  position  of  the 
other  face,  it  is  said  to  revolve  or  turn  through  the  dihe- 
dral angle,  and  the  greater  the  amount  of  turning  the 
greater  the  angle. 


LINES    AND    PLANES. 


267 


PROPOSITION  XXII.     THEOREM. 

372.  Two  dihedral  angles  are  equal  if  their  plane 
angles  are  equal. 


Let  CD  A  B  and  GHEFbe  two  dihedral  angles  whose 
plane  angles,  M  O  N  and  R  S  T,  are  equal. 

To  prove  that  the  dihedral  angles  CDAB  and  GHEF 
are  equal. 

SUG.  1.  Place  the  edge  A  D  upon  E  H,  so  that  Z. 
M  O  AT  coincides  with  Z  R  S  T. 

SUG.  2.  Where  will  the  plane  A  B  lie  with  respect  to 
the  plane  E  Ft  Where  will  A  C  lie  with  respect  to 
EG?  (Art.  339.) 

Therefore 

373.  Prove  the  converse  of  the  above  proposition; 
that  is,  if  two  dihedral  angles  are  equal  their  plane  angles 
are  equal. 


Ex.  265.  If  a  line  and  a  plane  are  parallel,  a  line 
drawn  from  any  point  in  the  plane  parallel  to  the  given 
line  will  lie  wholly  in  the  plane.  (See  Art.  359.) 


268 


SOLID    GEOMETRY. 


PROPOSITION  XXIII.     THEOREM. 

374.  The  ratio  of  two  dihedral  angles  equals  the 
ratio  of  their  plane  angles. 


Let  MN  and  G  K  be  two  dihedral  angles,  and  aOf 
and  p  Ht  their  plane  angles. 

dihedral  M  N          Z.  a  Of 


To  prove  that 


dihedral  GK 


CASE  I.      When  the  plane  angles  are  commensurable. 

SUG.  1.  Divide  the  plane  Z_$  by  their  common  unit  of 
measure.  Pass  planes  through  the  lines  of  division  and 
the  edges  of  the  dihedral  ^s. 

SUG.  2.  L,et  the  unit  of  measure  be  contained  m  times 
in  a  Of,  and  n  times  in  p  H  t.  Then  what  is  the  ra- 

tio 


SUG.  3.  How  do  the  smaller  dihedral  ^s  into  which 
the  given  dihedral  ^/s  are  divided  compare  ?  Why  ? 
How  many  small  dihedral  Z.§  in  the  given  dihedral 
Z.  M  N1  How  many  in  the  given  dihedral  /_  G  K1 

.     dihedral  AM  N 

SUG.  4.  What  is  the  ratio    ...    ,  —  7—7-,^  „    equal  to  ? 

dihedral  Z.  G  K 


LINES    AND    PLANES.  269 


SUG.  5.  Compare  answers  to  Sugs.  2  and  4. 

What  is  the  conclusion  in  Case  I  of  the  proposition  ? 

CASE  II.      When  the  angles  are  incommensurable. 

SUG.  Use  the  method  of  Art.  171,  Case  II,  and  give 
the  demonstration. 
Therefore 

375.  SCHOLIUM.  Since  dihedral  angles  are  propor- 
tional to  their  plane  angles,  they  are  said  to  be  measured 
by  their  plane  angles.  Thus  a  right  dihedral  angle  is 
one  whose  plane  angle  is  a  right  angle,  and  a  dihedral 
angle  of  27°  is  one  whose  plane  angle  is  an  angle  of  27°, 
etc. 

Compare  the  method  of  leading  up  to  the  measurement 
of  dihedral  angles  with  that  of  leading  up  to  the  meas- 
urement of  arcs. 


Ex.  266.  If  a  line  is  parallel  to  each  of  two  intersect- 
ing planes  it  is  parallel  to  their  intersection. 

Ex.  267.  Two  parallel  planes  are  everywhere  equi- 
distant. 

Ex.  268.  What  is  the  locus  of  a  point  at  a  given  dis- 
tance from  a  given  plane  ? 

Ex.  269.  A  straight  line  and  a  plane  perpendicular  to  the 
same  straight  line  are  parallel. 

Ex.  270.  If  a  straight  line  and  a  plane  are  parallel* 
any  line  parallel  to  the  line  is  parallel  to  the  plane  also. 

Ex.  271.  If  a  line  is  parallel  to  a  line  of  a  plane  it  is 
parallel  to  the  plane. 

Ex.  272.  Prove  that,  through  a  given  point  in  space, 
one  and  only  one  line  can  be  drawn  perpendicular  to  a 
given  line;  the  given  point  being  without  the  given  line.. 


270 


SOLID    GEOMETRY. 


PROPOSITION  XXIV.     THEOREM. 

376.  If  a  straight  line  is  perpendicular  to  a  plane, 
every  plane  containing  that  line  is  perpendicular  to 
the  plane. 


Let  A  JB  be  a  straight  line  perpendicular  to  the  plane 
M  Wf  and  let  C  D  be  any  plane  containing  the  line  A  B. 

To  prove  that  the  plane  CD  is  perpendicular  to  the 
plane  M  N. 

SUG.  1.  Through  B,  draw  a  line  B  E  in  the  plane 
•M  N  J_  to  B  D,  the  intersection  of  the  two  planes. 

SUG.  2.  The  Z  E  B  A  is  the  plane  angle  of  the  dihe- 
dral Z.NBD  A.  Why? 

SUG.  3.  How   many   degrees   in   Z  E  B  A  ?     Why  ? 

SUG.  4.  Then,  what  relation  does  the  plane  C  D  sus- 
tain to  the  plane  M  N?  (Art.  367.) 

Therefore 


Ex.  273.  Prove  that,  through  a  given  point  in  space, 
one,  and  only  one  line  can  be  drawn  parallel  to  a  given 
line. 

Ex.  274.  What  is  the  locus  of  the  foot  of  an  oblique 
line  10  in.  long  drawn  from  a  point  8  in.  from  a  plane  ? 


LINES    AND    PLANES. 


271 


PROPOSITION  XXV.     THEOREM. 

377.  If  two  planes  are  perpendicular  to  each  other, 
a  line  drawn  in  one  of  them  perpendicular  to  their  in- 
tersection is  perpendicular  to  the  other. 


Let  M  N  and  C  D  be  two  planes  perpendicular  to  each 
otfier,  intersecting  in  the  line  B  Z>,  and  let  A  B  be  a  line  in 
C  D  perpendicular  to  B  D. 

To  prove  that  A  B  is  perpendicular  to  the  plane  M  N. 

SUG.  1.   In  the  plane  M  N,  draw  B  E  _L  to  B  D,  at  B. 

SUG.  2.  The  Z.  A  B  E  is  the  plane  Z.  of  the  dihe- 
dral Z..  Why?  How  many  degrees  in  Z.  A  B  E  ?  Why? 

SUG.  3.  What  relation  does  A  B  sustain  to  the  plane 
AfN?  Why?  (Art.  346.) 

Therefore 

378.  COROLLARY  I.     If  two  planes  are  perpendicular 
to  each  other,  a  perpendicular  to  one  of  them  at  any  point 
of  their  intersection  lies  in  the  other. 

SuG.  1.  A  B,  in  the  plane  C  D,  is  _L  to  M  N  by  the 
proposition. 

SUG.  2.   How  many  _Ls  can  be  erected  to  M  N  at  B  ? 

379.  COROLLARY  II.     If  two  planes  are  perpendicular 
to  each  other,  a  perpendicular  to  one  of  them  from  any 
point  in  the  other  lies  in  the  other. 

SUG.  See  Sug.  1  of  Cor.  I,  and  Art.  343. 


272 


SOLID    GEOMETRY. 


PROPOSITION  XXVI.     THEOREM. 

380.  If  two  planes  are  perpendicular  to  a  third 
plane,  their  intersection  is  perpendicular  to  that  plane. 


Let  C  D  and  E  F  be  two  planes,  each  perpendicular  to 
the  plane  M  JV,  and  let  A  O  be  the  line  of  intersection  of 
C  D  and  E  F. 

To  prove  that  A  O  is  perpendicular  to  the  plane  M  N. 

SUG.  1.  From  the  point  O,  which  is  common  to  all 
three  planes,  erect  a  J_  to  the  plane  M  N. 

SUG.  2.  Where  will  this  J_  lie  with  respect  to  each  of 
the  planes  C  D  and  E  Ft  (Art.  378.) 

SUG.  3.  What  relation  does  the  J_  erected  sustain  to 
the  intersection  of  the  planes  C  D  and  E  Ft 

Therefore 


Ex.  275.  If  a  plane  intersects  two  parallel  planes  the 
alternate  interior  dihedral  angles  are  equal. 

Kx.  276.  If  one  plane  intersects  another,  the  sum  of 
the  two  adjacent  dihedral  angles  on  the  same  side  of 
either  plane  is  equal  to  two  right  dihedral  angles. 

Ex.  277.  If  a  plane  intersects  two  parallel  planes  the 
corresponding  dihedral  angles  are  equal. 


LINKS    AND    PLANES.  273 

PROPOSITION  XXVII.     THEOREM. 

381.  Through  a  given  straight  line,  one,  and  but 
one,  plane  can  be  passed  perpendicular  to  a  given 
plane. 


Af 


Let  ABbea  given  straight  line,  and  MNa  given  plane. 

To  prove  that  one,  and  but  one,  plane  can  be  passed 
through  A  B  perpendicular  to  M  N. 

SUG.  1.  From  some  point,  as  A,  in  A  B,  drop  a  _L 
A  Cto  the  plane  M  N. 

SUG.  2.  A  B  and  A  C  determine  a  plane  B  C.  Give 
auth. 

SUG.  3.  What  relation  does  the  plane  B  C  sustain  to 
the  plane  M  Nt  (Art.  376.) 

SUG.  4.   Can  a  plane  be  drawn  through  A  B  J_  to  M '  Nt 

SUG.  5.  How  many  planes  can  be  drawn  through  A  B 
J_  to  M  Nt  (Art.  339.) 

Therefore 


Ex.  278.  If  the  sum  of  two  adjacent  dihedral  angles  is 
equal  to  two  right  dihedral  angles,  their  exterior  faces 
are  in  the  same  plane. 

.     Ex.  279.  If  two  planes  intersect  each  other,  their  ver- 
tical dihedral  angles  are  equal. 
18— Geo. 


274  SOLID    GEOMETRY. 

PROPOSITION  XXVIII.     THKOREM. 

382,  Every  point  in  a  plane  which  bisects  a  dihe- 
dral angle  is  equidistant  from  the  faces  of  the  dihe- 
dral angle. 


Mite  a  plane  bisecting  the  dihedral  angle  formed 
by  the  planes  A  C  and  A  D ,  and  let  P  be  any  point  in  the 
plane  A  M,  and  P  E  and  P  F  perpendiculars  from  P  to 
the  faces  A  C  and  A  D  of  the  dihedral  angle. 

To  prove  that  P  E  and  P  F  are  equal. 

SuG.  1.  Pass  a  plane  through  the  lines  P  E  and  P  F 
and  let  it  intersect  the  planes  A  C,  A  M  and  A  D  in  the 
lines  O  E,  O  P  and  O  F. 

SuG.  2.  What  relation  does  the  plane  E  F  sustain  to 
each  of  the  planes  A  C  and  A  D  ?  Give  auth. 

SUG.  3.  What  relation  does  the  plane  E  F  sustain  to 
A  B,  the  edge  of  the  dihedral  Z.  ?  Give  auth. 

SUG.  4.   Compare  Zs  P  O  E  and  P  O  F.     (Art.  371.) 

SUG.  5.   Compare  As  P  O  E  and  POP.     Give  auth. 

SUG.  6.   Compare  the  lines  P  E  and  P  F. 

Therefore 


LINES    AND    PLANES. 


275 


383.  The  projection  of  a  point  upon  a  plane  is  the 
foot  of  the  perpendicular  from  the  point  to  the  plane. 

The  projection  of  a  line  /? 

upon  a  plane  is  the  line  in  the 
plane  which  contains  the  pro- 
jections of  all  the  points  of  the 
line.  In  the  figure  at  the 
right,  the  point  C  is  the  pro- 
jection  of  the  point  A,  and  the 


7 

/N 


line  C  D  is  the  projection  of  the  line  A  B. 


PROPOSITION  XXIX.     THEOREM. 

384.  The  projection  of  a  straight  line  upon  a  plane 
is  a  straight  line. 


Let  ABbea  given  straight  line  and  MN  a  given  plane. 

To  prove  that  the  projection  of  A  B  upon  the  pla?ie  M  N 
is  a  straight  line. 

Sue.  1.  Let  A  D  represent  the  plane  which  contains 
A  B  and  is  _L  to  the  plane  M  N.  This  plane  contains 
CD,  the  projection  of  A  B  upon  the  plane  M  N.  Why  ? 

SUG.  2.  The  two  planes  A  D  and  M  N  intersect  in 
what  kind  of  a  line  ? 

Therefore 


276  SOLID    GEOMETRY. 

PROPOSITION  XXX.     THEOREM. 

385.  The  angle  ivhich  a  straight  line  makes  with 
its  own  projection  upon  a  plane  is  the  least  angle  the 
line  malces  with  any  line  of  the  plane. 


Let  A  JB  represent  any  straight  line,  and  13  C  its  projec- 
tion upon  the  plane  MN. 

To  prove  that  the  angle  A  B  C  is  the  least  angle  the  line 
A  B  can  make  with  any  line  of  the  plane  M  N. 

SUG.  1.  Through  B,  draw  any  other  line  of  the  plane, 
as  B  D.  From  A,  drop  a  _L  A  C  to  the  plane  M  N. 
Ivay  off  B  D  =  B  C  and  draw  A  D. 

SUG.  2.  Compare  A  C  and  A  D  in  respect  to  length. 
Give  auth. 

SUG.  3.  Compare  /.ABC  with  Z.  A  B  D.  Give 
auth. 

Therefore 

386.  The  angle  which  a  line  makes  with  its  projec- 
tion on  a  plane  is  called  the  angle  of  the  line  and  the 
plane,  of  the  inclination  of  the  line  to  the  plane. 


LINES    AND    PLANES.  277 

POLYHEDRAL  ANGLES. 

387.  The  figure  formed  by  three  or  more  plaues  meet- 
ing at  a  common  point  is  called  a  polyhe- 
dral angle.  In  the  figure  at  the  right, 
A-B  CD  represents  a  polyhedral  angle. 
The  common  point  A  is  called  the  vertex 
of  the  angle;  A ' B,  A  C,  etc. ,  are  the  edges. 
The  portions  of  planes  intercepted  by  the 
edges;  as  B  A  C,  CAD,  etc.,  are  the  faces 
of  the  angle.  The  plane  angles  formed  by  the  edges,  as 
B  A  C,  CAD,  etc.,  are  the  face  angles  of  the  polyhe- 
dral angle. 

In  a  polyhedral  angle  each  pair  of  adjacent  faces  forms 
a  dihedral  angle;  as  B  C  A  D,  etc. 

A  polyhedral  angle  of  three  faces  is  called  a  trihedral 
angle,  one  of  four  faces  is  called  a  tetrahedral  angle, 
etc. 

Two  polyhedral  angles  are  equal  when  they  can  be 
made  to  coincide. 

Two  polyhedral  angles  are  symmetrical  when  the  parts 
of  one  one  are  equal  respectively  to  the  parts  of  the  other 
but  arranged  in  reverse  order. 


D  ^ 

^C' 

A-B  CD  and  A'-  B'  CD  are  symmetrical  if  Z.  B  A  C 
:  Z  B'  A'  C,  Z  C  A  D  =  Z.  C  A'  U  and  Z  B  A  D 
.  B'  A'  D'. 


278  SOLID   GEOMETRY. 

A  polyhedral  angle  cannot  be  made  to  coincide  with 
its  symmetrical  polyhedral  angle. 

Note. —  The  two  hands,  or  the  two  sides  of  the  face,  of  the  human 
body  will  serve  to  illustrate  symmetrical  solids.  To  convince  one 
that  the  two  hands  are  symmetrical  rather  than  equal,  it  might  be 
suggested  to  try  to  put  the  left  glove  on  the  right  hand.  See  defini- 
tion of  equal  polyhedral  angles. 

PROPOSITION  XXXI.    THEOREM. 

388.  The  swm  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle. 


B 

"Let  ABC  represent  a  trihedral  angle  in  which  each  of 
the  face  angles  DAB  and  B  A  C  is  smaller  than  the  face 
angle  D  A  C. 

To  prove  that  the  sum  of  the  angles  DAB  and  B  A  C 
is  greater  than  the  angle  D  A  C. 

SUG.  1.  In  the  face  DA  C  draw  A  M,  making  Z  DA  M 
=  Z.DAB. 

SuG.  2.  Cut  the  edges  by  a  plane  D  B  C,  so  that  A  M 
shall  =  A  B. 

SUG.  3.  Compare  &$DAB  and  DAM.  Compare  D  M 
and  B  D.  Give  auth.  Compare  B  CandMC.  Give  auth. 

SuG.  4.   Compare  /_BAC  with  Z  MA  C.    Give  auth. 

SUG.  5.  Compare  the  sum  of  Zs  DAB  and  B  A  C 
with  Z.  DA  C. 

Therefore 


LINES    AND   PLANES. 


279 


PROPOSITION  XXXII.     THEOREM. 

389.   The  sui) i  of  the  face  angles  of  any  convex  poly- 
hrdral  angle  is  less  than  four  right  angles. 


Let  A-BCDEF  represent  a  convex  polyhedral  angle. 

To  prove  that  the  sum  of  the  face  angles  B  A  C,  C  A  Dy 
etc. ,  is  less  than  four  right  angles. 

SUG.  1.  Pass  a  plane  cutting  the  edges  of  the  polyhe- 
dral Z  in  the  points  B,  C,  D,  E  and  F. 

SUG.  2.  Connect  O,  any  point  within  the  polygon 
B  C  D  E  F,  with  each  of  the  vertices. 

SUG.  3.  In  the  trihedral  Z  B  compare  the  sum  of  the 
face  Zs  A  B  C  and  A  B  F  with  the  face  Z  C  B  F. 

SUG.  4.  In  the  trihedral  Z  C,  compare  the  sum  of  the 
face  Zs  A  C  B  and  A  CD  with  the  face  Z.  B  C D. 

SUG.  5.  Compare  the  sum  of  the  base  Zs  of  the  As 
whose  vertices  are  at  A,  with  the  sum  of  the  base  Zs  of 
the  As  whose  vertices  are  at  O. 

SUG.  6.  Compare  the  sum  of  the  vertical  Zs  at  A  with 
the  sum  of  the  vertical  Zs  at  O. 

SUG.  7.  Compare  the  sum  of  the  Zs  at  A  with  four 
rt.  Zs. 

Therefore • 


280 


SOLID    GEOMETRY. 


PROPOSITION   XXXIII.     THEOREM. 

390.  If  two  trihedral  angles  have  the  three  face 
angles  of  one  equal  respectively  to  the  three  face 
angles  of  the  other t  the  corresponding  dihedral  angles 
are  equal.  , 


Let  A-BCD  and  A'-B'C'D'  represent  two  trihedral 
angles  whose  face  angles  are  equal ;  viz.,  B  A  C  =  B'A '  €' , 
CAD=C'A'D  and  B  A  D  =  B'  A  D'. 

To  prove  that  the  corresponding  dihedral  angles  CB  A  D 
and  C'  B'  A'  D ',  etc.,  are  equal. 

SUG.  1.  Pass  planes  BCD  and  B'  C  D ',  making  the 
edges  A  B,  A  C,  A  D,  A'  B',  A'  C1  and  A'  D1  all  equal. 

SUG.  2.  In  A  B  and  A'  B'  take  points  M  and  M'  so 
that  A  M  =  A'  M9,  and  at  M  and  M'  pass  planes  J_  to 
A  B  and  A'  B'  respectively. 

SUG.  3.  The  plane  M  O  N  must  intersect  B  C  and  B  D, 
or  these  lines  extended,  and  the  plane  M'  (7  N1  must  in- 
tersect B'  C'  and  B'  Z7,  or  these  lines  extended.  Why  ? 

SUG.  4.  What  relation  do  the  lines  MO  and  M  N  sus- 
tain to  A  B  ?  Why  ?  What  relation  do  the  lines  M1  a 
and  M  N'  sustain  to  A  B'  ?  Why  ? 


LINES    AND    PLANES.  281 

SUG.  5.  Compare  As  B  A  C  and  B'  A'  Cr.  Compare 
Z.*A  £  CandA'  B'  C'. 

SUG.  6.  Compare  As  B  MO  and  B'  Af  <7.  (Art.  101.) 
Compare  B  O  with  B'  a. 

SUG.  7.  In  a  similar  manner  compare  As  B  M  N  and 
^  Af'  A77.  Compare  ^  JV  with  B'  N' . 

SUG.  8.  Compare  As  C  B  D  and  C'  B'  D '.  Compare 
Zs  C  B  D  and  C'  B'  D' . 

SUG.  9.  Compare  As  O  B  N  and  a  B'  W .  Compare 
ON  and  0  N1. 

SUG.  10.  Compare  As  O  M  N  and  <7  M'  N'.  Com- 
pare Zs  O  M  N  and  (7  A/"  Nf. 

Therefore 

391.  SCHOLIUM.     When  the  parts  of  A-B  CD  are  ar- 
ranged in  the  same  order  as  the  respectively  equal  parts 
of  A'-  B'  C'  D'  the  two  dihedral   angles  are  equal,  for 
they  may  then  be  superposed  (i.  e.,  placed  one  upon  an- 
other so  as  to  coincide  throughout),  but  when  the  parts 
of  one  are  arranged  in  the  reverse  order  from  the  respect- 
ively equal  parts  of  the  other,  the  two  trihedral  angles 
are  symmetrical. 

Note. —  The  student  should  attempt  to  superpose  A' -  B'  C'  D' 
upon  A-B  CD.  If  A'  B'  be  placed  upon  A  B  what  must  be  consid- 
ered in  order  that  A '  C'  may  fall  upon  A  Cl  When  the  face  B'  A'  C1 
is  thus  made  to  coincide  with  the  face  B  A  C  the  edges  A1  D'  and 
A  D  may  fall  upon  the  same  or  opposite  side  of  the  plane  B  A  C.  If 
the  two  edges  A'  D'  and  A  D  fall  upon  the  same  side  of  B  A  C  the 
two  trihedral  angles  are  equal,  but  if  they  fall  upon  opposite  sides  of 
BAG  the  two  trihedral  angles  are  symmetrical. 

392.  An    isosceles   trihedral   angle   is   a   trihedral 
angle  two  of  whose  face  angles  are  equal. 


282 


SOLID    GEOMETRY. 


PROPOSITION  XXXIV.     THEOREM. 

393.  Two  symmetrical  isosceles  trihedral  angles 
are  equal  in  all  respects. 


Let  A  B  CD  and  A'  B'  C'  D'  represent  two  symmet- 
rical isosceles  trihedral  angles  in  which  the  face  angles 
B  A  C  and  CAD  are  equal  •  also  the  face  angles  B'  A'  C' 
and  C'  A'  D'  are  equal. 

To  prove  that  the  trihedral  angles  A-BCD  andA'-B'C'H 
are  equal  in  all  respects. 

SUG.  1.  Pass  planes  BCD  and  B'  C  D' ,  cutting  the 
edges  of  the  two  trihedral  Zs,  so  that  A  B,  A  C,  A  D, 
A'  B',  A'  C  and  A1  D>  are  all  equal. 

SUG.  2.   Compare  the  face  Zs  B  A  C  and  D'  A'  C'. 

SuG.  3.  Compare  the  face  Zs  C  A  D  and  C'  A'  B' . 

SUG.  4.  Superpose  A'-B'  C  D1  upon  A-BCD,  placing 
A'  upon  A,  D'  upon  B  and  C'  upon  C.  Why  is  this 
possible  ?  Where  will  the  face  C  A'  B'  fall  with  respect 
to  the  face  CA  D!  Why? 

SUG.  5.  Where  will  the  line  A '  B'  fall  ?     Why  ? 

SUG.  6.  Where  will  the  face  D'  A '  B'  lie  ? 

Therefore 


PROPOSITIONS   IN   CHAPTER  VI. 


PROPOSITION  I. 

A  plane  may  be  passed  through  any  straight  line,  but  the  line  does 
not  determine  the  plane. 

PROPOSITION  II. 

A.  plane  is  determined: 

I.     By  a  straight  line  and  a  point  without  the  line. 
II.     By  two  interseciing  straight  lines. 

III.  By  two  parallel  straight  lines. 

IV.  By  three  points  not  in  a  straight  line. 

PROPOSITION  III. 
The  intersection  of  two  planes  is  a  straight  line. 

PROPOSITION  IV. 

From  a  given  point  without  a  plane,  one,  and  only  one,  perpen- 
dicular can  be  dropped  to  the  plane,  and  the  perpendicular  is  the 
shortest  line  from  the  point  to  the  plane. 

PROPOSITION  V. 

At  a  given  point  in  a  plane,  one  perpendicular,  and  only  one,  can 
be  erected  to  the  plane. 

PROPOSITION  VI. 

If,  from  any  point  in  the  perpendicular  to  a  plane,  oblique  lines 
be  drawn,  those  which  meet  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular  are  equal;  and,  of  two  unequal  oblique  lines, 
that  which  meets  the  plane  at  the  greater  distance  from  the  foot  of 
the  perpendicular  is  greater. 

PROPOSITION  VII. 

If  a  straight  line  is  perpendicular  to  two  lines  of  a  plane  at  their 
point  of  intersection,  it  is  perpendicular  to  the  plane. 


SOLID    GEOMETRY. 


PROPOSITION  VIII. 

All  the  perpendiculars  to  a  given  line  at  the  same  point  lie  in  the 
•same  plane,  and  that  plane  is  perpendicular  to  the  given  line. 

PROPOSITION  IX. 

If,  from  the  foot  of  a  perpendicular  to  a  plane,  a  line  be  drawn 
perpendicular  to  any  line  of  the  plane,  and,  from  the  point  of  inter- 
section, a  line  be  drawn  to  any  point  of  the  perpendicular,  the  las* 
line  is  perpendicular  to  the  line  of  the  plane. 

PROPOSITION  X. 

If  one  of  two  parallel  straight  lines  is  perpendicular  to  a  plane,  the 
other  is  perpendicular  to  the  same  plane. 

PROPOSITION  XI. 
If  two  lines  are  perpendicular  to  the  same  plane,  they  are  parallel. 

PROPOSITION  XII. 

If  two  straight  lines  are  each  parallel  to  a  third  straight  line,  they 
are  parallel  to  each  other. 

PROPOSITION  XIII. 

Every  plane  containing  one,  and  only  one,  of  two  parallel  lines,  is 
parallel  to  the  other  line. 

PROPOSITION  XIV. 

If  a  line  and  a  plane  are  parallel,  the  intersection  of  the  plane  with 
any  plane  containing  the  line  is  parallel  to  the  line. 

PROPOSITION  XV. 
Planes  perpendicular  to  the  same  straight  line  are  parallel. 

PROPOSITION  XVI. 

The  intersections  of  two  parallel  planes  with  a  third  plane  are  par- 
allel lines. 

PROPOSITION  XVII. 

If  two  angles  not  in  the  same  plane  have  their  sides  parallel  and 
lying  in  the  same  direction  from  their  vertices,  they  are  equal  and 
their  planes  are  parallel. 


LINES    AND    PLANES.  285 


PROPOSITION  XVIII. 

A  straight  line  perpendicular  to  one  of  two  parallel  planes  is  per- 
pendicular to  the  other  also. 

PROPOSITION  XIX. 

If  three  parallel  planes  are  intersected  by  two  lines,  the  segments 
of  the  lines  are  proportional. 

PROPOSITION  XX. 
All  plane  angles  of  the  same  dihedral  angle  are  equal. 

PROPOSITION  XXI. 

If  a  plane  be  passed  perpendicular  to  the  edge  of  a  dihedral  angle, 
the  lines  of  intersection  wiih  the  faces  form  the  plane  angle  of  the 
dihedral  angle. 

PROPOSITION  XXII. 
Two  dihedral  angles  are  equal  if  their  plane  angles  are  equal. 

PROPOSITION  XXIII. 

The  ratio  of  two  dihedral  angles  equals  the  ratio  of  their  plane 
angles. 

PROPOSITION  XXIV. 

If  a  straight  line  is  perpendicular  to  a  plane,  every  plane  contain- 
ing that  line  is  perpendicular  to  the  plane. 

PROPOSITION   XXV. 

If  two  planes  are  perpendicular  to  each  other,  a  line  drawn  in  one 
of  them  perpendicular  to  their  intersection  is  perpendicular  to  the 
other. 

PROPOSITION  XXVI. 

If  two  planes  are  perpendicular  to  a  third  plane,  their  intersection 
is  perpendicular  to  that  plane. 

PROPOSITION  XXVII. 

Through  a  given  straight  line,  one,  and  but  one,  plane  can  be 
passed  perpendicular  to  a  given  plane. 


286  SOLID    GEOMETRY. 


PROPOSITION  XXVIII. 

Every  point  in  a  plane  which  bisects  a  dihedral  angle  is  equidis- 
tant from  the  faces  of  the  dihedral  angle. 

PROPOSITION  XXIX. 
The  projection  of  a  straight  line  upon  a  plane  is  a  straight  line. 

PROPOSITION  XXX. 

The  angle  which  a  straight  line  makes  with  its  own  projection  upon 
a  plane  is  the  least  angle  the  line  makes  with  any  line  of  the  plane. 

PROPOSITION  XXXI. 

The  sum  of  any  two  face  angles  of  a  trihedral  angle  is  greater  than 
the  third  face  angle. 

PROPOSITION  XXXII. 

The  sum  of  the  face  angles  of  any  convex  polyhedral  angle  is  less 
than  four  right  angles. 

PROPOSITION  XXXIII. 

If  two  trihedral  angles  have  the  three  face  angles  of  one  equal  re- 
spectively to  the  three  face  angles  of  the  other,  the  corresponding  di- 
hedral angles  are  equal. 

PROPOSITION  XXXIV. 

Two  symmetrical  isosceles  trihedral  angles  are  equal  in  all  re- 
spects. 


CHAPTER  VII. 
POLYHEDRONS. 


•  DEFINITIONS. 

394.  A  polyhedron  is  a  geometric  solid  bounded  by 
planes. 

The  bounding  planes  of  a  polyhedron  are  its  faces,  the 
lines  in  which  the  faces  intersect  are  the  edges,  and  the 
points  in  which  the  edges  intersect  are  the  vertices  of 
the  polyhedron. 

Any  face  designated  may  be  considered  the  base  of  the 
polyhedron. 

395.  A  straight  line  joining  any  two  vertices  not  in 
the  same  face,  is  called  a  diagonal  of  the  polyhedron. 

396.  The  intersection  of  a  polyhedron  and  a  plane  is 
called  a  plane  section  of  the  polyhedron. 

397.  Polyhedrons  are  classified  according  to  the  num- 
ber of  their  faces. 

A  polyhedral  angle  requires  at  least  three  planes  to 
meet  at  its  vertex,  and  to  completely  inclose  space  re- 
quires at  least  one  more  plane;  hence  a  polyhedron  can- 
not have  less  than  four  faces. 

398.  A  polyhedron  of  four  faces  is  called  a  tetrahe- 
dron; one  of  five  faces,  a  pentahedron;  one  of  six  faces, 
a  hexahedron;  one  of  eight  faces,  an  octahedron;  one 
of  ten  faces,  a  decahedron;  one  of  twelve  faces,  a  do- 
decahedron; one  of  twenty  faces,  an  icosahedron,  etc. 


288 


SOLID    GEOMETRY. 


ICOSAHEDRON. 


DODECAHEDRON. 


OCTAHEDRON*. 


HEXAHEDRON. 


TETRAHEDRON. 


399.  A  convex  polyhedron  is   one   in  which  every 
possible  plane  section  is  a  convex  polygon. 

PRISMS. 

400.  A  prism  is  a  polyhedron  two  of  whose  faces  are 
polygons  which  are  equal  in  all  respects  and  are  in  par- 
allel planes,  and  whose  remaining  faces  are  parallelo- 
grams. 

401.  The  bases  of  a  prism  are  the  equal  faces  in  the 
parallel  planes.     The  lateral  faces  are  the  remaining 
faces  of  the  prism. 

402.  The  basal  edges  of  a  prism  are  the  intersections 
of  the  lateral  faces  with  the  bases.     The  lateral  edges 
are  the  intersections  of  the  lateral  faces. 

403.  A  right  section  of  a  prism  is  a  section  whose 
plane  is  perpendicular  to  the  lateral  edges  of  the  prism. 


POLYHEDRONS. 


289 


404.  The  altitude  of  a  prism  is  the  perpendicular  dis- 
tance between  its  bases. 

405.  Prisms  are  classified  as  triangular,  quadrangu- 
lar, etc.,  according  as  their  bases  are  triangles,  quadri- 
laterals, etc. 


\D 


*\ 


RIGHT  PRISM. 


OBLIQUE  PRISM. 


406.  A  right  prism  is  one  in  which  the  lateral  edges 
are  perpendicular  to  the  bases,  and  an  oblique  prism  is 
one  in  which  the  edges  are  oblique  to  the  bases. 

407.  A  regular  prism  is  a  right  prism  whose  bases 
are  regular  polygons. 


Ex.  280.  Parallel  lines  in- 
tersecting the  same  plane 
make  equal  angles  with  it. 

Let  A  B  and  C  D  represent 
||  lines,  and  B  O  and  D  S 
their  respective  projections  on 
the  plane  M  N. 

SUG.  1.  Drop  J_s  A  O  and  C  S  to  the  plane  M  N. 

SUG.  2.  Compare  /LBA  O  with  ZDCS.    (Art.  362.) 

SUG.  3.  Compare  Z  B  with  Z  D. 
19— Geo. 


290  SOLID    GEOMETRY. 

PROPOSITION  I.     THEOREM. 

408.  The  lateral  edges  of  a  prism  are  equal  and 
parallel,  and  make  equal  angles  with  the  plane  of 
either  'base. 


Let  A  Ft  B  H,  etc.,  represent  the  lateral  edges  of  a 
prism. 

To  prove  that  A  F,  B  H,  etc. ,  are  equal  and  parallel, 
and  that  they  make  equal  angles  with  the  plane  of  either 
base. 

SuG.  1.  What  kind  of  a  figure  is  the  lateral  face  A  HI 
Then,  what  relation  must  A  F  and  B  H  sustain  to  each 
other  ? 

SuG.  2.  What  relation  must  any  two  lateral  edges  sus- 
tain to  each  other  ? 

SUG.  3.  Compare  the  Z.§  made  by  any  two  lateral 
edges  with  either  base.  Give  auth.  (See  Ex.  280.) 

Therefore 


Ex.  281.  If  a  plane  intersects  two  parallel  planes  the 
interior  dihedral  angles  on  the  same  side  of  the  cutting 
plane  are  supplements  of  each  other. 


POLYHEDRONS.  291 


PROPOSITION  II.     THEOREM. 

409.  Sections  of  a  prism  made  by  parallel  planes 
are  polygons  which  are  equal  in  all  respects. 


Let  A  C  and  ac  be  sections  of  the  prism  F  K9  made  by 
parallel  planes. 

To  brove  that  the  polygons  A  C  and  a  c  are  equal  in  all 
respects. 

SUG.  1.  What  relation  does  A  E  sustain  to  a  £?  Why  ? 
(Art.  361.) 

SUG.  2.  What  relation  does  E  D  sustain  to  e  d,  D  C 
to  d  c,  etc.? 

SUG.  3.  Compare  Zs  A  ED  and  a  e  d\  also  Zs  ED  C 
and  edc,  etc.  Give  auth. 

Therefore 


Hx.  282.  Bvery  section  of  a  prism  made  by  a  plane 
parallel  to  a  lateral  edge  is  a  parallelogram. 

Ex.  283.  If  two  planes  are  cut  by  a  third  plane  so  that 
the  alternate  interior  dihedral  angles  are  equal,  and  the 
edges  of  the  dihedral  angles  thus  formed  are  parallel,  the 
two  planes  are  parallel. 


292  SOLID    GEOMETRY. 

PROPOSITION  III.    THEOREM. 

410.  The  lateral  area  of  a  prism  equals  the  product 
of  a  lateral  edge  by  the  perimeter  of  a  right  section  of 
the  prism. 


Let  C  D  be  a  lateral  edge,  and  MP  a  right  section  of 
the  prism  A  D. 

To  prove  that  the  area  of  the  lateral  faces  of  the  prism 
equals  C  D  'multiplied  by  the  perimeter  of  M  P. 

SUG.  1.  What  relation  does  P  O  sustain  to  C  D  ? 
Why  ?  What  relation  does  any  side  of  the  right  sec- 
tion sustain  to  a  lateral  edge  which  meets  that  side  ? 

Sue.  2.  What  is  the  area  of  the  face  C  E,  in  terms  of 
POsiud  CD? 

SUG.  3.  Express  the  area  of  any  lateral  face. 

SUG.  4.  Add  the  areas  of  the  several  lateral  faces,  re- 
membering that  the  lateral  edges  are  equal. 

Therefore  


Ex.  284.  If  two  planes  are  cut  by  a  third  plane  so  that 
the  corresponding  dihedral  angles  are  equal  and  the 
edges  of  the  dihedral  angles  formed  are  parallel,  the  two 
planes  are  parallel. 


POLYHEDRONS.  293 


Note.  —  The  algebraic  expression  ab+ac  +  ad,  may  be  reduced,  by 
factoring,  to  the  form  a  (b-\-c-\-d).  Compare  with  the  former  ex- 
pression the  indicated  sum  of  the  areas  of  the  lateral  faces  and  re- 
duce it,  by  the  algebraic  process  of  factoring,  to  the  product  of  a 
lateral  edge  by  the  perimeter  of  a  right  section  of  the  prism. 


Ex.  285.  In  any  trihedral  angle  the  three  planes  passed 
through  the  edges  and  the  bisectors  of  the  respectively 
opposite  face  angles  intersect  in  a  straight  line. 

Ex.  286.  In  any  trihedral  angle  the  three  planes  bi- 
secting the  dihedral  angles  intersect  in  a  straight  line. 

Ex.  287.  Determine  a 

*  >¥ 

point  in  a  plane  the  differ-  M  S 

ence  of  whose  distances  from  p  ..•'  \ 

two  given  points  on  oppo- 
site sides  of  the  plane  is  the 
maximum.  x  ;  ** 

SUG.   Drop   a    _L    to   the  SJ^ 

plane  from  one  of  the  points,  as  B,  and  extend  it  to  M, 
an  equal  distance  on  the  other  side  of  the  plane.  Con- 
nect A  and  M,  and  extend  the  line  to  meet  the  plane  at  O. 
Prove  that  O  is  the  required  point. 

Ex.  288.  Determine  a 

point  in  a  plane  the  sum  of  /Jx  / 

whose  distances  from  two 
given  points  on  the  same 
side  of  the  plane  is  the  mini- 
mum. 


SUG.   Drop   a  _L  to   the  'M 

plane  from  one  of  the  points  A,  and  extend  to  M,  an 
equal  distance  beyond  the  plane.  Connect  M,  the  ex- 
tremity of  the  JL,  with  the  other  point  B. 

Prove  that  O,  the  point  in  which  MB  intersects  the 
plane,  is  the  required  point. 


294 


SOLID    GEOMETRY. 


PROPOSITION  IV.     THEOREM. 

411.  Two  prism>s  are  equal  in  all  respects  if  the 
three  faces  about  a  trihedral  angle  of  one  are  respect- 
ively equal  in  all  respects  to  the  three  faces  about  a 
trihedral  angle  of  the  other,  and  these  faces  are  sim- 
ilarly placed  in  the  two  figures. 


Let  A  H  and  A' H'  be  two  prisms  in  which  the  three 
faces  A  Z>,  A  G  and  A  K  are  respectively  equal  in  all  re- 
spects to  tfie  three  faces  A'D'9  A'G'  and  A  K' ,  and  are 
similarly  placed. 

To  prove  that  the  prisms  A  H  and  A'  H1  are  equal  in 
all  respects. 

SUG.  1.  Compare  the  trihedral  ^.s  A  and  A '.  (Arts. 
390  and  391.) 

SUG.  2.  Apply  the  prism  A'  H'  to  the  prism  A  H,  so 
that  the  face  A'  Dr  coincides  with  the  face  A  D.  Why 
can  this  be  done? 

SUG.  3.  In  what  plane  will  the  face  A'  G'  fall  ?    Why  ? 

SUG.  4.  In  what  plane  will  the  face  A'  K1  fall  ?    Why  ? 

SUG.  5.  Where  will  the  line  A'  F'  fall  ?     Why  ? 

SUG.  6.  Where  will  the  point  F'  fall  ?  Where  will  the 
point  G'  fall  ?  Why  ?  Where  will  the  point  K'  fall  ? 
Why? 


POLYHEDRONS.  295 


SUG.  7.  Since  F't  G  and  K'  fall  upon  F,  G  and  ^re- 
spectively, in  what  plane  will  the  face  Fr  I'  fall  ?  Why  ? 
(Art.  339.) 

SUG.  8.  Since  B' ,  G  and  C'  fall  upon  B,  G  and  C  re- 
spectively, in  what  plane  will  the  face  B'  ff  fall  ?  Where 
will  the  line  C'  fT  fall  ? 

SUG.  9.  Similarly,  where  will  the  line  D'  I'  fall  ? 
Where  will  the  points  f?  and  /'  fall  ? 

SUG.  10.   Compare  the  prism  A'ff  with  the  prism  A  H. 

Therefore 

412.  COROLLARY.     Two  right  prisms  are  equal  in  all 
respects  if  their  altitudes  are  equal,  and  the  bases  of  one 
are  equal  in  all  respects  to  the  bases  of  the  other. 

413.  A  truncated  prism  is  a  portion  of  a  prism  in- 
cluded between  a  base  and  a  section  of  the  prism  by  a 
plane  not  parallel  to  the  base. 

PROPOSITION  V.     THEOREM. 

414.  Two  truncated  prisms  are  equal  in  all  respects 
if  the  three  faces  about  a  trihedral  angle  of  one  are 
respectively  equal  in  all  respects  to  the  three  faces 
about  a  trihedral  angle  of  the  other,  and  these  faces 
are  similarly  placed  in  the  two  figures. 

SUG.  Use  the  method  of  Prop.  IV. 


Ex.  289.  If  a  plane  be  passed  through  the  edge  of  a 
dihedral  angle  in  such  a  manner  that  every  point  in  the 
plane  is  equidistant  from  the  faces  of  the  dihedral  angle, 
the  plane  bisects  the  dihedral  angle. 


SOLID    GEOMETRY. 


PROPOSITION  VI.     THEOREM. 

415.  An  oblique  prism  is  equal  in  magnitude  to  a 
right  prism  whose  base  is  a  right  section  of  the  oblique 
prism  and  whose  altitude  is  a  lateral  edge  of  the 
oblique  prism. 


Let  B  K  be  an  oblique  prism  and  A'B  C' D' E'  a  right 
section. 

To  prove  that  the  prism  B  K  is  equal  in  magnitude  to  a 
right  prism  having  A'  B1  C'  D'  E'  for  a  base  and  an  alti- 
tude equal  to  E  K. 

SUG.  1.  Extend  the  lateral  edge  E  K  to  K' ',  making 
E  K'  =  E  K.  Through  K'  pass  a  plane  K'  G'  \\  to  the 
plane  E!  B'.  Extend  the  other  lateral  edges  to  meet  the 
plane  K'  G  in  the  points  /',  H' ,  G  and  F' . 

SUG.  2.  Compare  /'  Dr  with  K'  E'  and  hence  with 
K  E.  Compare  /  D  with  K  E,  Hence,  compare  /'  D' 
with  ID.  Compare  //'  with  D  D' . 

SUG.  3.  Compare  K'  I'  with  E'D';  also  K'  K  with  E'E. 

SUG.  4.  Compare  the  trapezoid  K'  I  with  the  trape- 
zoid  E'  D. 

SUG.  5.  Similarly  compare  the  trapezoid  K1  F  with 
the  trapezoid  E'  A. 


POLYHEDRONS.  297 


SUG.  6.  Compare  Z  I'  K'  F'  with  Z.  D'  E'  A'.  Com- 
pare the  polygon  K'  G'  with  the  polygon  E' B' . 

SUG.  7.  Compare  the  trihedral  Z  I'  K'  K  F  with  the 
trihedral  Z  D'E' E  A'. 

SUG.  8.  Compare  the  truncated  prism  K'  G  with  the 
truncated  prism  E' B. 

SUG.  9.  Compare  the  right  prism  K'  B'  with  the 
oblique  prism  K  B  in  respect  to  magnitude. 

Therefore 


Ex.  290.  Prove  that  the  lateral  area  of  a  right  prism 
is  less  than  the  lateral  area  of  any  oblique  prism  having 
the  same  base  and  an  equal  altitude. 

PARAU<EI,OPIPEDS. 

416.  A  parallelopipe'd  is  a  prism  whose  bases  are 

parallelograms. 

417.  A  right  parallelepiped  is  a  parallelepiped  whose 
lateral  edges  are  perpendicular  to  the  bases. 

418.  A  rectangular  parallelepiped  is  a  right  paral- 
lelepiped whose  bases  are  rectangles.- 

419.  A  cube  is  a  rectangular  parallelepiped  whose 
faces  are  all  squares. 

420.  The  volume  of  a  polyhedron  is  its  ratio  to  some 
selected  unit  of  measure,  joined  to  the  name  of  that  unit. 
For  example  if  a  cubic  inch  is  contained  twenty-five 
times  in  a  given  polyhedron  the  volume  of  the  polyhe- 
dron is  twenty-five  cubic  inches. 

The  unit  of  measure  for  volume  is  a  cube  whose  edge 
is  a  given  linear  unit. 


298  SOLID    GEOiMETRY. 


PROPOSITION  VII.     THEOREM. 

421.   Opposite  faces  of  a  parallelopiped  are  par- 
allelograms which  are  equal  in  all  respects. 


C 
Let  A  Gbea  parallelopiped. 

To  prove  that  two  opposite  faces,  as  A  H  and  B  G,  are 
equal  in  all  respects. 

SUG.  1.  What  kind  of  a  polygon  is  A  H?  What  kind 
is#£?  Why? 

SUG.  2.  Compare  lines  A  D  and  B  C  ;  A  E  and  B  F. 
Compare  Z!s  D  A  R  and  C  B  F. 

SUG.  3.  Then  how  must  the  face  A  H  compare  with 
the  face  B  G  ?  Why  ? 

SUG.  4.   Compare  the  faces  A  F  and  D  G. 

Therefore 


Ex.  291.  If  any  plane  be  passed  through  either  diag- 
onal of  a  parallelogram,  the  perpendiculars  to  this  plane 
from  the  extremities  of  the  other  diagonal  are  equal. 

Ex.  292.  Having  given  a  fixed  straight  line,  and  two 
points  not  in  the  line,  find  a  point  in  the  fixed  line 
equally  distant  from  the  two  fixed  points. 


POLYHEDRONS.  299 


PROPOSITION  VIII.     THEOREM. 

422.  A  plane  passed  through  the  diagonally  oppo- 
site edges  of  a  parallelepiped  divides  it  into  two  tri- 
angular prisms  which  are  equal  in  magnitude. 

A 
^ 

D 


G 

Let  A  G  be  a  paraU<'lo))i/><><f,  and  D  B  F  H  a  plane 
imsswl  through  thediitgomiUu  opposite  edges  1)11  and  />/'. 

To  prove  that  the  triangular  prisms,  EFH-A  ana 
GFH-C,  into  which  the  pnsm  is  divided  are  equal  in 
volume. 

SUG.  1.  Pass  a  right  section  M  N  O  P  through  the  par- 
allelopiped. 

SUG.  2.   Compare  A  M  N  O  with  A  M  P  O. 

SUG.  3.  Compare  the  triangular  prism  whose  base  is 
M  N  O  and  whose  altitude  is  H  D  with  the  triangular 
prism  whose  base  is  M  P  O  and  whose  altitude  is  H  D, 

SUG.  4.  How  is  the  triangular  prism  whose  base  is 
M  N  O  and  whose  altitude  is  H  D  related  to  the  triangu- 
lar prism  EP  H-A  ? 

SUG.  5.  How  is  the  triangular  prism  whose  base  is 
M  P  O  and  whose  altitude  is  H  D  related  to  the  triangular 
prism  G  F H-Ct 

SUG.  6.  Compare  triangular  prism  E  H F-A  with  tri- 
angular prism  G FH-C. 

Therefore 


300 


SOLID    GEOMETRY. 


PROPOSITION  IX.     THEOREM. 

423.  Any  parallelepiped  is  equal  in  volume  to  a 
rectangular  parallelopiped  having  an  equal  altitude 
and  a  base  equal  in  area. 


Let  E  Cite  a  given  parallelopiped. 

To  prove  that  E  C  is  equal  in  volume  to  a  rectangular 
parallelopiped  having  an  equal  altitude  and  a  base  equal  in 
area. 

SUG.  1.  Extend  the  edges  A  D,  B  C,  F  G  and  E  H. 
Take  E'  If  =  Effand  pass  the  right  sections  E'  B'  and 
If  C  JL  to  E'  H'. 

SUG.  2.  Compare  the  parallelepipeds  h  C  and  E'  C'  in 
volume.  (Art.  415.) 

SUG.  3.  Extend  the  edges  A' B',  D'  C',  E1  F  and 
If  G'.  Take  S  R  =  H'  G'  and  pass  the  right  sections 
6"  K  and  R  L  _L  to  S  R. 

SUG.  4.  Compare  the  parallelepipeds  O  M  and  E'  C' 
in  volume.  (Art.  415.) 

SUG.  5  Compare  the  parallelepipeds  E  C  and  O  M  in 
volume. 


POLYHEDRONS.  301 


SUG.  6.  Since  the  plane  R  L  was  passed  J_  to  5  RY 
2.  e.,  _L  to  K  L,  what  does  Z.  K  L  M  equal  ?  Hence, 
what  kind  of  a  figure  Js  L  N1 

SUG.  7.  Since  the  plane  L  R  is  JL  to  K  L,  what  does 
Z  A-ZP equal? 

SUG.  8.  The  plane  E  B'  was  passed  _L  to  E  IT. 
Hence,  how  is  the  plane  O  L  related  to  L  M"?  Hence, 
what  does  Z.  ML  /'equal? 

SUG.  9.  Since  P  L  is  J_  to  L  K  and  L  M,  how  is  P  L 
related  to  the  plane  L  Nt  Hence  what  kind  of  a  par- 
allelepiped is  K Rt 

SUG.  10.  Compare  the  area  of  K  M  with  the.  area  of 
A'  C',  also  the  area  of  A'  €'  with  the  area  of  A  C. 
Hence,  compare  the  area  of  K  M  with  the  area  of  A  C. 

SUG.  11.  Compare  the  altitudes  of  these  three  par- 
allelepipeds. 

Therefore 


Ex.  293.  A  plane  perpendicular  to  one  edge  of  a  trihe- 
dral angle  intersects  the  faces  in  lines  which  form  a  right 
triangle. 

Hx.  294.  The  four  diagonals  of  a  rectangular  parallel- 
epiped are  equal  to  one  another. 

Ex.  295.  Any  straight  line  drawn  through  the  middle 
point  of  any  diagonal  of  a  parallelepiped  terminating  in 
two  opposite  faces  is  bisected  at  that  point. 

Ex.  296.  If  from  any  point  in  space  perpendiculars  are 
drawn  to  the  lateral  faces  of  a  prism,  these  perpendiculars 
are  all  in  the  same  plane. 

Ex  297.  If  from  any  point  in  space  perpendiculars  are 
drawn  to  the  lateral  edges  of  a  prism,  these  perpendicu- 
lars are  all  in  the  same  plane. 


302 


SOLID    GEOMETRY. 


PROPOSITION  X.     THEOREM. 

424.   Two    rectangular   parallelopipeds   having 
equal  bases  have  the  same  ratio  as  their  altitudes. 


H 


Let  A  &  and  E  H  represent  two  rectangular  parallelo 
pipeds  whose  bases  A  C  and  E  G  are  equal,  and  whose  al 
titudes  are  AB  and  E  F  respectively  . 

41       A  D        AB 
To  prove  that 


CASE  I.  When  the  altitudes  A  B  and  E  F  are  com- 
mensurable. 

SUG.  1.  Lay  off  a  common  measure  upon  the  altitudes 
A  B  and  E  F,  and  let  this  common  measure  be  contained 
m  times  in  A  B  and  n  times  in  E  F. 

A  B 
SUG.  2.  What  does  the  ratio  -=-    equal? 


SUG.  3.  Pass  right  sections  of  the  parallelopipeds 
through  the  points  of  division  of  the  altitudes  A  B  and 
E  F,  thus  dividing  the  given  parallelopipeds  into  smaller 
parallelopipeds.  Compare  these  smaller  parallelopipeds 
in  respect  to  volume.  (Art.  412.) 

SUG.  4.  Compare  the  number  of  small  parallelopipeds 
into  which  the  given  parallelopipeds  are  divided  with  the 
number  of  segments  into  which  the  altitudes  are  divided. 


A  D 
SUG.  5.  What  does  the  ratio    - 


equal  ? 


AD 
SUG.  6.  Compare  the  ratios  - 


and 


A  B 


^  «  . 

Jb  Jrl  £L,  r 

SUG.  7.  What  is  the  conclusion  when  the  altitudes  are 
commensurable  ? 


POLYHEDRONS. 


303 


CASE  II.      When  the  altitudes  are  incommensurable. 
r> 

sS\      s\ 

B 

/ 

F 


SUG.  1.  If  A  B  and  E  F  are  incommensurable,  lay  off 
any  measure  of  R  /''upon  A  B  as  many  times  as  possible. 
There  will  be  a  remainder,  as  KB,  less  than  the  meas- 
ure. Why  ? 

SUG.  2.  Through  K  pass  a  right  section  K  L  ot  the 
parallelepiped  A  D. 

A  L  A  K 

SUG.  3.   Compare  the  ratios  -^r-^  and  -~-~-.  (Case  L) 

£,  Jri  £L  r 

SUG.  4.  If  a  line  shorter  than  KB  be  taken  as  a  unit 
of  measure  of  E  F,  the  remainder  will  be  some  line  MB, 
which  is  less  than  KB.  Why  ? 

SUG.  5.  By  making  the  unit  of  measure  of  E  F  smaller 
and  smaller  continually,  the  remainder  MB  is  made  to 
decrease  indefinitely.  Why? 

SUG.  6.  What  relation  always  exists  between  the  ra- 

A  N        ,  A  M      TT7U    _ 
tios  -r;  and   --.     Why? 


A  N  A  M 

SUG.  7.  Are  -^-77  and  -=,---  variables  ?  Why?  What 
Jz,  rl  £,  r 

are  their  respective  limits  ? 

A  N          A  M 

SUG.  8.  Compare  the  limits  of  -frjj-  and     _  _  .    Give 

LL  ri  h,  r 

auth. 

SUG.  9.  What  is  the  conclusion  when  the  altitudes  are 
incommensurable  ? 

Therefore  - 


304: 


SOLID    GEOMETRY. 


425.  COROLLARY.     If  two  rectangular  parallelepipeds 
have  two  edges  of  one  equal  respectively  to  two  edges  of 
the  other,  the  ratio  of  the  parallelepipeds  equals  the  ra- 
tio of  the  third  edges. 

PROPOSITION  XI.    THEOREM. 

426.  The  number  of  units  of  volume  in  a  rectangu- 
lar parallelepiped  is  equal  to  the  product  of  the  num- 
ber of  linear  units  in  the  edges  meeting  at  any  vertex. 


Let  A  represent  any  rectangular  parallelopiped,  and 
let  a,  b  and  c  represent  three  edges  meeting  at  a  verteoo. 
Let  V  be  the  unit  of  measure  for  volume,  the  edge  of  U  be- 
ing the  linear  unit  u.  Let  ^  =  m,  —  =  n,  and  —  —  r;  m 
representing  the  number  of  linear  units  in  a,  n  in  b,  and 

r  in  c. 

^ 

To  prove  that  -=j-  =  mxnxr. 

SUG.  1.  Construct  the  parallelepiped  B,  two  of  whose 
edges  meeting  at  a  vertex  are  b  and  c  and  the  third  edge 
is  u,  the  edge  of  the  unit  U. 

SUG.  2.  Construct  another  parallelepiped  C,  whose 
edges  meeting  at  a  vertex  are  c  (the  remaining  edge  of  A}, 
u  and  u. 


POLYHEDRONS.  305 


Sue.  3.  What  does  the  ratio  -^  equal?     Give  auth. 

n 

SUG.  4.   What  does  the  ratio  -=  equal?     Give  auth. 

SUG.  5.  What  does  jj-  equal  ?     Give  auth. 
Therefore 

427.  SCHOLIUM  I.     In  the  applications  of  this  theorem 
the  three  edges  must  be  expressed  in  terms  of  the  same 
unit,  and  the  unit  of  volume  must  be  a  cube  whose  edge 
is  the  linear  unit. 

428.  SCHOLIUM  II.     By  comparison  of  the  theorem 
with   the   definition   of  volume   (Art.   420),   it   will   be 
observed  that  the  volume  of  a  rectangular  parallelepiped 
is  equal  to  the  product  of  the  measures  of  three  edges 
meeting  at  any  vertex  joined  to  the  name  of  the  unit  of 
measure  for  volume. 

429.  SCHOLIUM  III.     The  expression  ''the  product  of 
the  three  dimensions,"  is  a  common  abbreviation  for  the 
expression,  "the  product  of  the  measures  of  three  edges 
meeting  at  any  vertex  joined  to  the  name  of  the  unit  of 
measure  for  volume." 

The  product  of  three  lines  must  not  be  interpreted  in 
any  other  sense  than  that  just  stated.  With  this  inter- 
pretation, Prop.  XI  is  usually  stated  as  follows:  The 
volume  of  a  rectangular  parallelepiped  is  equal  to  the 
product  ol  its  three  dimensions. 

430.  SCHOLIUM  IV.     When  each  dimension  of  the 
rectangular  parallelepiped  is  di- 
visible by  the  linear  unit,  which 

is  the  edge  of  the  unit  of  volume, 
the  truth  of  the  theorem  may  be 
shown  by  dividing  the  parallele- 
piped into  cubes,  each  equal  to 
the  unit  of  measure.  This  method 
is  usually  employed  in  arithmetic. 
20— Geo. 


306  SOLID  GEOMETRY. 

431.  SCHOLIUM  V.     If  the  three  edges  of  a  rectangu- 
lar parallelepiped  meeting  at  any  vertex  are  equal,  the 
volume  is  equal  to  the  third  power  of  the  linear  unit, 
and  hence  the  third  power  of  a  number  is  called  the  cube 
of  the  number. 

PROPOSITION  XII.    THKOREM. 

432.  The  volume  of  any  parallelopiped  is  equal  to 
the  product  of  the  area  of  its  base  by  its  altitude. 

SUG.  See  propositions  IX  and  XI. 

433.  COROLLARY  I.    If  two  parallelepipeds  have  bases 
which  are  equal  in  area,  the  ratio  of  their  volumes  equals 
the  ratio  of  their  altitudes. 

SUG.  1.  Let  V,  B  and  A  represent  respectively  the 
volume,  the  area  of  the  base  and  the  altitude  of  one  par- 
allelopiped, and  F',  B'  and  A'  the  volume,  area  of  the 
base  and  altitude  of  the  other  parallelopiped.  Then  by 
the  theorem  V  =  By,  A  and  V  =  B'xA'. 

SUG.  2.  If  now  B  =  B',  what  is  the  ratio  -^  equal  to  ? 

434.  COROLLARY  II.      If  two  parallelepipeds   have 
equal  altitudes,  the  ratio  of  their  volumes  equals  the  ra- 
tio of  the  areas  of  their  bases. 

SUG.  As  in  Cor.  I,   V=  BxA,  and  V  =  B'xA'.     If 

V 
A  =  A'  what  is  the  ratio  -y^  equal  to  ? 


Ex.  298.  In  a  given  plane  find  a  point  which  is  equally 
distant  from  the  vertices  of  a  given  triangle. 


POLYHEDRONS. 


307 


PROPOSITION  XIII.    THEOREM. 

435.  The  volume  of  a  triangular  prism  is  equal  to 
the  product  of  the  area  of  its  base  by  its  altitude. 


Let  A  B  C-F  represent  a  triangular  prism. 

To  prove  that  the  volume  of  A  B  C-F  is  equal  to  the  area 
of  A  B  C  multiplied  by  the  altitude  of  the  prism-. 

SUG.  1.  Draw  A  D  ||  to  B  C,  and  C  D  ||  to  B  A.  Also 
draw  E  H  ||  to  F  G,  and  G  H  \\  to  F  E.  Connect  D 
with  H. 

SUG.  2.  What  kind  of  a  figure  is  A  B  CD-F1     Why  ? 

SUG.  3.  What  is  the  volume  of  A  B  C D-F  equal  to? 
Why? 

SUG.  4.  Compare  the  prism  A  B  C-F  with  the  figure 
ABC  D-F  in  respect  to  volume.  Give  auth.  (See 
Prop.  VIII.) 

SUG.  5.  Compare  the  bases  ABC  and  A  B  CD  of  the 
figures  A  B  C-F  and  ABC  D-F  respectively .  Also  com- 
pare the  altitudes  of  these  two  figures. 

SUG.  6.  What  is  the  volume  of  the  prism  A  B  C-F  in 
terms  of  the  area  of  its  base  and  its  altitude  ? 

Therefore 


308 


SOLID    GEOMETRY. 


PROPOSITION  XIV.     THEOREM. 

436.  The  volume  of  any  prism  is  equal  to  the  prod- 
uct of  the  area  of  its  base  by  its  altitude. 

A 

B 


Let  B  K  represent  any  prism,  and  A  B  CD  E  its  base. 

To  prove  that  the  volume  of  B  K  is  equal  to  the  area  of 
ABODE  multiplied  by  the  altitude  of  the  prism. 

SUG.  1.  Through  any  lateral  edge,  as  A  /%  pass  diag- 
onal planes  A  H,  etc. 

SUG.  2.  Into  what  kind  of  figures  is^the  given  prism 
divided  by  these  planes  ? 

SUG.  3.  What  is  the  volume  of  A  B  C-G  equal  to? 
The  volume  of  A  CD- HI  etc.  Give  auth. 

SUG.  4.  Express  the  sum  of  these  volumes  in  the  sim- 
plest form. 

SUG.  5.  What  is  the  volume  of  B  K  in  terms  of  its  base 
and  altitude  ? 

Therefore 

437.  COROLLARY  I.  If  two  prisms  have  bases  which 
are  equal  in  area,  their  volumes  have  the  same  ratio  as 
their  altitudes. 


POLYHEDRONS.  309 


438.  COROLLARY  II.     If  two  prisms  have  equal  alti- 
tudes their  volumes  have  the  same  ratio  as  the  areas  of 
their  bases. 

PYRAMIDS. 

439.  A  pyramid  is  a  polyhedron  all  but  one  of  whose 
faces  meet  in  the  same  point.  A 

The  point  in  which  all  the  faces  but 
one  meet  is  called  the  vertex. 

The  face  which  does  not  pass  through 
the  vertex  is  the  base. 

Tiie  faces  which  meet  at  the  vertex 
are  called  lateral  faces. 

The  edges  formed  by  the  intersections  of  the  lateral 
faces  are  called  the  lateral  edges. 

The  edges  formed  by  the  intersections  of  the  base  with 
the  lateral  faces  are  called  the  basal  edges. 

440.  The  altitude  of  a  pyramid  is  the  perpendicular 
distance  from  the  vertex  to  the  plane  of  the  base. 

441.  A  pyramid  is  called  triangular,  quadrangular, 
pentagonal,  etc.,    according  as  its  base  is  a   triangle, 
quadrilateral,  pentagon,  etc. 

442.  A  regular  pyramid  is  one  whose  base  is  a  regu- 
lar polygon  and  whose  vertex  is  in  a  per- 
pendicular to  the  base  erected  at  its  mid- 
dle point. 

443.  The  slant  height  of  a  regular 
pyramid  is  the  perpendicular  from  the 
vertex  of  the  pyramid  to  any  basal  edge. 


310 


SOLID  GEOMETRY. 


444.  A  truncated  pyramid  is  the  portion  of  a  pyra- 
mid included  between  the  base  and  a  plane  cutting  all 
its  lateral  edges. 

445.  The  frustum  of  a  pyramid  is  a  truncate  pyra- 
mid in  which  the  cutting  plane  is  par- 
allel to  the  base.     The  section   of  the 

pyramid  made  by  the  cutting  plane  is 
the  upper  base  of  the  frustum. 

446.  The  altitude  of  the  frustum  of  a 
pyramid  is  the  perpendicular  distance 
between  its  bases. 

PROPOSITION  XV.    THEOREM. 

447.  If  a  pyramid  be  cut  by  a  vlane  parallel  to  the 
base: 

I.  The  edges  and  altitude  are  divided  proportion- 
ally. 

II.  The  section  is  a  polygon  similar  to  the  base. 


Let  A-G  HKL.M  represent  a  pyramid,  BCDEF  a 
section  made  by  the  plane  S  T  par  all  el  to  the  base,  A  O  the 
altitude,  and  P  the  point  in  which  A  O  intersects  the 
plane  S  T. 


POLYHEDRONS.  311 


T       _  ,,   ,AB        AC        AD      4          AP 

I.  To  prove  that  _  ==  _  =  _,  etc.,  -  -^  . 

SUG.  1.  Through  A,  pass  a  plane  N  R  \\  to  the  plane 

GL. 

A  B     AC      AP 
SUG.  2.  Compare  -^,    ^^,    -^Q>  etc.     (See  Art. 

364.) 

II.  To  prove  that  B  CD  E  F  is  similar  to  G  HKL  M. 

SUG.  1.  Compare  As  A  B  F  and  A  G  Min  respect  to 
form.  Also  As  A  F  E  and  A  ML,  etc. 

BF     FE      ED 
SUG.  2.  Compare  the  ratios  -^,    ^    z^)  etc. 

Give  auth. 

SUG.  3.  Compare  Z.  B  F  E  with  Z.  G  M  L.  Also 
/.FED  with  Z.MLK,  etc. 

SUG.  4.  Then,  what  relation  does  the  polygon  /?  ^ 
sustain  to  the  polygon  G  L  ? 

SUG.  5.  Notice  the  conclusions  of  parts  I  and  II. 

Therefore  - 

ADDITIONAL  HELPS. 

In  II,  Sug.  1,  ,#/MS  U  to  GM,  FE  is  ||  to  ML.  Why? 
(Art.  361.) 

DC* 

In  II,  Sug.  2.  To  compare  the  ratios      -      and 


y 

compare  each  with  the  ratio       ..  . 

448.  COROLLARY  I.     The  areas  of  parallel  sections  of 
a  pyramid  are  proportional  to  the  squares  of  the  distances 
of  the  cutting  planes  from  the  vertex.     (See  Art.  268.) 

449.  COROLLARY  II.     In  pyramids  whose  bases  are 
equal  in  area  and  whose  altitudes  are  equal,  sections  at 
equal  distances  from  the  vertices  are  equal  in  area. 


312 


SOLID    GEOMETRY. 


PROPOSITION  XVI.     THEOREM. 

450.   Two  triangular  pyramids  having  equal  alti- 
tudes and  bases  equal  in  areaf  are  equal  in  volume. 


B 


Let  S-ABC  and  S'-A'B'C'  represent  two  triangular 
pyramids  having  equal  altitudes  and  bases  ABC  and 
A'B'C'  equal  in  area. 

To  prove  that  S-ABC  and  S'-A'B'C  are  equal  in 
volume. 

SUG.  1.  Divide  the  altitude  A  R  of  the  pyramids  into 
equal  parts,  and  through  the  points  of  division,  M,  N 
and  (9,  pass  planes  ||  to  the  bases,  thus  making  sections 
of  the  pyramids.  Upon  each  section  of  S-A  B  C  as  up- 
per base  construct  a  prism  whose  altitude  is  equal  to  the 
_L  distance  between  the  sections  and  whose  lateral  edges 
are  ||  to  the  edge  S  A  of  the  pyramid  S-ABC.  Simi- 
larly, upon  each  section  of  S'-A'B'C'  as  upper  base, 
construct  a  prism  whose  altitude  is  equal  to  the  _L  dis- 
tance between  the  sections  and  whose  lateral  edges  are  || 
to  the  edge  S' A '  of  the  pyramid  S'-A'B'C. 


POLYHEDRONS.  313 


Sue.  2.  Compare  the  areas  of  the  sections  DBF  and 
D'E'F'.  Giveauth. 

SUG.  3.  Compare  the  prisms  D  E F-A  and  D'  E'  F'-A ' ; 
also  the  two  prisms  adjacent  to  them;  also  the  next  two, 
etc.  Give  auth. 

SUG.  4,  Compare  the  sum  of  the  prisms  in  S-A  B  C 
with  the  sum  of  the  prisms  in  S'-A'B'C'. 

SUG.  5.  Is  the  answer  to  Sug.  4  true,  whatever  the 
number  of  parts  into  which  the  altitude  is  divided  ? 

SUG.  6.  If  the  number  of  parts  into  which  the  altitude 
is  divided  be  increased  continually  the  sum  of  the  prisms 
in  A- B  C  D  will  be  a  variable;  also  the  sum  of  the 
prisms  in  A'-B'  C  D'  will  be  a  variable.  Why? 

SUG.  7.  How  are  these  two  variables  related  ? 

SUG.  8.  What  are  their  respective  limits  ? 

Therefore 


Ex.  299.  The  square  of  a  diagonal  of  a  rectangular 
parallelepiped  equals  the  sum  of  the  squares  of  the  three 
dimensions. 

Ex.  300.  Find  the  length  of  a  diagonal  of  a  rectangu- 
lar parallelepiped  1  ft.  long,  4  in.  wide  and  3  ft.  high. 

Ex.  301.  Find  the  lateral  area  of  a  regular  triangular 
prism  whose  basal  edges  are  each  4  ft.  and  whose  lateral 
edges  are  each  2  yds. 

Ex.  302.  Find  the  total  area  of  a  regular  triangular 
prism  whose  basal  edges  are  each  4  ft.,  and  whose  lateral 
edges  are  each  8  ft. 

Ex.  303.  Find  the  volume  of  a  regular  hexagonal 
prism  whose  basal  edges  are  each  2  ft. ,  and  whose  lateral 
edges  are  each  2  yds. 


314: 


SOLID    GEOMETRY. 


PROPOSITION  XVII.     THEOREM. 

451.  The  volume  of  a  triangular  pyramid  is  one 
third  the  volume  of  a  triangular  prism  having  the 
same  base  and  altitude. 


Let  A-B  CD  represent  a  triangular  pyramid. 

To  prove  that  the  volume  of  A-B  CD  is  one  third  the 
volume  of  a  prism  having  BCD  for  a  base  and  an  altitude 
equal  to  the  altitude  of  the  pyramid. 

SUG.  1.  Through  B,  draw  BE  ||  and  equal  to  C  A, 
and  through  D  draw  D  F  ||  and  equal  to  C  A.  Con- 
nect E  with  A,  E  with  F,  and  A  with  F,  thus  forming 
the  A  E  A  F. 

SUG.  2.  What  is  the  figure  EA  F-B  CD  thus  formed  ? 
Why?  (Arts.  362  and  400.) 

SuG.  3.  If,  from  the  figure  E  A  F-B  CD,  the  pyramid 
A-B  CD  be  removed  there  remains  the  figure  A-B  D  FE. 
What  is  this  figure  which  remains  ?  What  kind  of  a 
quadrilateral  is  B  D  F  E  ?  Why  ? 

SUG.  4.  Compare  the  two  pyramids  A-E  B  D  and 
A-EFD.  (Prop.  XVI.) 

SUG.  5.  Read  the  pyramid  A-EFD  as  D-E  A  F,  i.  e., 
consider  E  A  F  as  the  base  and  D  as  the  vertex,  and 
then  compare  D-E  A  .Fwith  A-B  C  D.  (Prop.  XVI.) 


POLYHEDRONS.  315 


SUG.  6.  What  part  of  the  whole  figure  is  A-B  CD  ? 
Therefore 

452.  COROLLARY.     The  volume  of  a  triangular  pyra- 
mid is  equal  to  one  third  of  the  product  of  the  area  of  its 
base  by  its  altitude. 

PROPOSITION  XVIII.     THEOREM. 

453.  The  volume  of  any  pyramid  is  equal  to  one 
third  of  the  product  of  the  area  of  its  base  by  its  alti- 
tude. 

O 


Let  O-A BCDE represent  a  pyramid. 

To  prove  that  the  volume  of  O-A  B  CD  E  is  equal  to  one 
third  the  area  of  the  base  ABODE  multiplied  by  its  alti- 
tude. 

SUG.  1.  Through  an  edge,  as  O  A,  pass  all  possible 
diagonal  planes. 

SUG.  2.  What  is  the  figure  O-ABC1  the  figure  O-A  CD  ? 
etc. 

SUG.  3.  What  is  the  volume  of  O-A  B  C  ?  of  O-A  CD  ? 
etc.  Give  auth. 

SUG.  4.  Express,  in  its  simplest  form,  the  sum  of  the 
volumes  of  the  figures  O-A  B  C,  O-A  CD,  etc. 

SUG.  5.  What  is  the  volume  of  the  pyramid  O-ABCDE  ? 

Therefore 


316 


SOLID   GEOMETRY. 


454.  COROLLARY  I.     If   two    pyramids    have    bases 
which  are  equal  in  area,  their  volumes  have  the  same 
ratio  as  their  altitudes. 

455.  COROLLARY  II.     If  two   pyramids   have  equal 
altitudes  their  volumes  have  the  same  ratio  as  the  areas 
of  their  bases. 

PROPOSITION  XIX.     THEOREM. 

456.  The  volume  of  the  frustum  of  a  triangular 
pyramid  is  equal  to  the  sum  of  the  volumes  of  three 
triangular  pyramids  whose  common  altitude  is  the 
altitude  of  the  frustum  and  whose  bases  are  respect- 
ively the  upper  base  of  the  frustum,  the  lower  base  of 
the  frustum,  and  a  mean  proportional  between  the 
bases  of  the  frustum. 

B 


Let  A  B  C-D  E  F  represent  the  frustum  of  a  triangular 
pyramid. 

To  prove  that  the  volume  of  A  B  C-D  E  F  is  equal  to  the 
sum  of  the  volumes  of  three  pyramids,  each  having  the  same 
altitude  as  the  frustum,  a?id  whose  bases  are  respectively  ABC, 
DEF,  and  a  mean  proportional  between  ABC  and  D  E  F. 


POLYHEDRONS.  317 


SUG.  1.  Through  the  edge  B  C  and  the  vertex  Dy  pass 
a  plane  BCD. 

SUG.  2.  If  A  B  C  be  considered  as  the  base  of  the  pyra- 
mid D-A  B  C,  how  does  the  altitude  of  this  pyramid  com- 
pare with  the  altitude  of  the  frustum  ?  Compare  the  pyr- 
amid D-ABC  with  one  of  the  pyramids  mentioned  in 
the  theorem. 

SUG.  3.  Pass  the  plane  C  D  E>  cutting  off  the  pyramid 
C-DE  F.  Compare  this  pyramid  with  one  of  the  pyra- 
mids mentioned  in  the  theorem. 

SUG.  4.  The  pyramid  D-C  B  E  still  remains.  Draw 
C  M  ||  to  B  E,  and  connect  D  with  M. 

SUG.  5.  Compare  the  pyramids  D-C  B  E  and  D-C  ME. 
(Art.  450.) 

SUG.  6.  Consider  C  the  vertex  and  D  M  E  the  base. 
How  does  the  altitude  of  C-D  M  E  compare  with  the 
altitude  of  the  frustum  ? 

SUG.  7.  Draw  MN\\  to  C  A. 


n    &DME         DE 
SUG.  9.  --  ^FVF-ET  =  -TFF-      Why? 
A  N  M  E         NE 


D  F 

SUG.  10.  Compare  the  ratios    ,  .  „    and     ._  „  .    Give 

ME  NE 


auth. 


SUG.  11.  Compare  the  ratios  and 

A  NME 


SUG.  12.  What  relation  does  £±  D  M  E  sustain  to 
AsDFjE.ind  N  M  El 

SUG.  13.  Compare  A  NME  with  &  A  C  B. 

SUG.  14.  Then,  what  relation  does  £±  D  M  E  sustain 
to  the  bases  D  F  E  and  A  C  B~t  Compare  the  pyramid 
C-D  M  E  with  one  of  the  pyramids  mentioned  in  the 
theorem. 

SUG.  15.  Review  Sugs.  2,  3  and  14. 

Therefore  - 


318 


SOLID    GEOMETRY. 


PROPOSITION  XX.     THEOREM. 

457.  The  volume  of  the  frustum  of  any  pyramid  is 
equal  to  the  sum  of  the  volumes  of  three  pyramids 
whose  common  altitude  is  the  altitude  of  the  frustum 
and  whose  bases  are  respectively  the  upper  base  of  the 
frustumt  the  lower  base  of  the  frustum,  and  a  mean 
proportional  between  the  bases  of  the  frustum. 

o 


Let  L  B  represent  tlie  frustum  of  any  pyramid. 

To  prove  that  the  volume  of  L  B  is  equal  to  the  sum  of 
the  volumes  of  three  pyramids  each  having  the  same  alti- 
tude as  L,  B,  and  whose  bases  are  respectively  the  upper 
base  L  G  of  the  frustum,  the  lower  base  E  B  of  the  frus- 
tum, and  a  mean  proportional  between  the  bases  L,  G  and 
EB. 

SUG.  1.  Let  the  lateral  edges  of  the  frustum  L  B  be 
extended  until  they  meet  at  the  vertex  O,  thus  forming 
the  pyramid  O-A  B  CDE.  Let  V-M N P  be  a  triangu- 
lar pyramid  whose  altitude  is  equal  to  the  altitude  of 
O-A  B  CDE,  and  whose  base  MNP  is  equal  in  area 
to  the  base  A  B  C  D  E.  Let  R  S  T  be  a  section  of  the 


POLYHEDRONS.  319 


triangular  pyramid  ||  to  the  base  MNP,  and  the  same 
distance  from  the  vertex  V  that  the  plane  L  G  is  from 
the  vertex  O. 

SUG.  2.  Compare  the  sections  L  G  and  R  S  T  in  re- 
spect to  area.  Give  auth.  (Art.  449.) 

Sue.  3.  Compare  the  pyramids  O-A  B  C  D  E  and 
V-MNPm  respect  to  volume. 

SUG.  4.  Compare  the  pyramids  O-FGHKL  and 
V-S  T  R  in  respect  to  volume. 

SUG.  5.  Compare  the  frustum  L  B  with  the  frustum 
RS  T-M  N  P  in  respect  to  volume.  Compare  the  alti- 
tudes of  these  two  frustums. 

SUG.  6.  What  is  the  volume  of  the  frustum  RST-MNP 
equal  to? 

SUG.  7.  What  then  is  the  volume  of  the  frustum  L  B 
eqaul  to  ? 

Therefore  - 

458.  COROLLARY.  If  the  volume  of  the  frustum  of  a 
pyramid  be  represented  by  F,  the  altitude  by  H,  the 
area  of  the  lower  base  by  B,  and  the  area  of  the  upper 
base  by  b,  the  truth  of  the  theorem  may  be  expressed  by 
the  formula 


Similarly  the  volume  of  a  prism  is  given  by  the  formula 

V=HB, 
and  the  volume  of  a  pyramid  by  the  formula 


Ex.  304.  Find  the  volume  of  a  regular  triangular  prism 
each  edge  of  which  is  4  ft. 

Ex.  305.  Find  the  lateral  area  of  a  regular  pentagonal 
prism  each  edge  of  which  is  3  in. 


320 


SOLID    GEOMETRY. 


PROPOSITION  XXI.     THEOREM. 

459.  The  lateral  area  of  a  regular  pyramid  is 
equal  to  one  half  the  product  of  the  perimeter  of  the 
base  by  the  slant  height. 


Let  O-A  BCDE  represent  a  regular  pyramid,  and  O  M 
its  slant  height. 

To  prove  that  the  lateral  area  of  the  pyramid  equals  one 
half  the  perimeter  of  the  base  multiplied  by  O  M. 

SUG.  1.  What  is  the  area  of  A  O  A  E  ?  of  A  O  E  D  ? 
of  A  O  D  C?  etc. 

SUG.  2.  What  is  the  lateral  area  of  the  pyramid  equal  to  ? 

Therefore 


Ex.  306.  Theorem.  The  volume  of  a  truncated  trian- 
gular prism  is  equal  to  the  sum  of 
the  volumes  of  three  pyramids 
whose  common  base  is  the  base  of 
the  prism  and  whose  vertices  are 
the  vertices  of  the  inclined  section. 

I^et  ABC-DEF  represent  a  trun-J 
cated  triangular  prism. 

SUG.  1.    Draw  the  lines  A  D,  A  Fy  B  D,  BE,  C  R 
and  CF. 


POLYHEDRONS. 


321 


SUG.  2.  Consider  the  truncated  prism  separated  into 
the  three  figures  B-D E F,  B-A  ED  and  B-A  CD. 

SUG.  3.  Compare  £-D£Fwiih  one  of  the  pyramids 
mentioned  in  the  theorem. 

SUG.  4.  Consider  B-A  E  D  as  D-AEB.  Show  that 
D-A  EB  =  D-A  EF.  Consider  D-A  EF  as  A-DEF 
and  compare  with  one  ot  the  pyramids  mentioned  in  the 
theorem. 

SUG.  5.  Compare  B-A  C D  with  B-ECD.  Consider 
B-E  CD  as  E-B  CD.  Compare  E-B  CD  with  E-FCD. 
Consider  E-FCD  as  C-D  E F,  and  compare  with  one  of 
the  pyramids  mentioned  in  the  theorem. 

PROPOSITION  XXII.     THEOREM. 

460.  The  lateral  area  of  the  frustum  of  a  regular 
pyramid  equals  the  product  of  the  slant  height  by  one 
half  the  sum  of  the  perimeters  of  the  two  bases. 

C. 


u 

Let  G  E  represent  the  frustum  of  a  regular  pyramid 
and  NM  its  slant  height. 

To  prove  that  the  lateral  area  of  the  frustum  G  E  equals 
M  N  multiplied  by  one  half  the  sum  of  the  perimeters  of  the 
two  bases. 

SUG.  1.  What  is  the  area  of  the  trapezoid  G  Ct  of  the 
trapezoid  H  D  ?  of  K  E  ?  etc. 
SUG.  2.  What  is  the  lateral  area  of  the  frustum? 

Therefore 

21— Geo. 


322  SOLID   GEOMETRY. 

461.  A  regular  polyhedron  is  a  polyhedron  whose 
faces  are  all  equal  regular  polygons  and  whose  polyhe- 
dral angles  are  all  equal. 

PROPOSITION  XXIII.     THEOREM . 

462.  Only  five  regular  convex    polyhedrons  are 
possible. 

SUG.  1.  The  regular  polygon  which  has  the  smallest 
number  of  sides  is  the  equilateral  A. 

SUG.  2.  How  many  degrees  are  there  in  an  Z.  of  an 
equilateral  A  ? 

SUG.  3.  Is  it  possible  for  three  equilateral  As  to  meet 
so  as  to  form  a  polyhedral  Z.  ?     (Art.  389.) 

SUG.  4.  What  is  the  least 
number  of  equilateral  As  that 
can  be  used  to  enclose  space  ? 

There  is  a  regular  polyhedron 
having  four  equilateral  As  for 
faces.  It  is  called  a  regular 
tetrahedron. 

SUG.  5.  Is  it  possible  for  four  equilateral  As  to  meet 
at  a  vertex  so  as  to  form  a  polyhedral  Z  ?     Why  ? 

SUG.  6.  How   many  equilateral   As   are   required   to 
completely  enclose  space  if  four  As  meet 
at  each  vertex  ? 

There  is  a  regular  polyhedron  formed 
by  eight  equal  equilateral  As.  It  is 
called  a  regular  octahedron. 

SUG.  7.  Is  it  possible  for  five  equilateral 
As  to  meet  at  a  vertex  so  as  to  form  a  polyhedral  /L  ? 
Why? 

SUG.  8.  How  many  equilateral  As  are  required  to  com- 
pletely enclose  space  if  five  As  meet  at  each  vertex  ? 


POLYHEDRONS.  323 


Perhaps  the  question  in  Sug.  8  cannot  be  answered 
without  the  aid  of  a  figure,  but  by  means  of  a  figure  con- 
structed of  cardboard,  as  suggested  below,  it  is  easily 
seen  that  there  is  a  regular  polyhedron 
formed  by  twenty  equal  equilateral  As. 
It  is  called  a  regular  icosahedron. 

SUG.  9.  Is  it  possible  for  six  equi- 
lateral As  to  meet  at  a  vertex  so  as  to 
form  a  polyhedral  Z.  ?  Why  ? 

SUG.  10.  Are  any  other  regular  con- 
vex polyhedrons  possible  whose  faces  are  equilateral  As  ? 

SUG.  11.  How  many  degrees  in  each  Z.  of  a  square? 

SUG.  12.   Is  it  possible  for  three  squares  to  meet  at  a 
vertex  so  as  to  form  a  polyhedral  4L  ?     Why  ? 

SUG.  13.   How   many  squares   are   required   to   com- 
pletely enclose  space  if  three  squares  meet  at  each  vertex  ? 

There  is  a  regular  polyhedron 
formed  by  six  equal  squares.  It  is 
called  a  regular  hexahedron  or  cube. 

SUG.  14.  Is  it  possible  for  more 
than  three  squares  to  meet  at  a  vertex 
so  as  to  form  a  polyhedral  Z  ?  Why  ? 


Are  any  other  regular  convex  polyhedrons  possible 
whose  faces  are  squares  ? 

SUG.  15.  How  many  degrees  in  each  Z  of  a  regular 
pentagon  ? 

SUG.  16.  Is  it  possible  for  three  regular  pentagons  to 
meet  at  a  vertex  so  as  to  form  a  polyhedral  Z  ?  V/hy  ? 

SUG.  17.  With  three  regular  pentagons  meeting  at 
each  vertex,  how  many  will  be  required  to  completely 
enclose  space  ? 


SOLID    GEOMETRY. 


The  question  in  Sug.  17  may  be  too  difficult  without 
the  aid  of  a  figure,  but  by  means  of  a  fig- 
ure constructed  of  cardboard  as  suggested 
below,  it  is  easily  seen  that  there  is  a 
regular  convex  polyhedron  formed  by 
twelve  regular  pentagons.  It  is  called  a 
dodecahedron. 

SUG.  18.  Is  it  possible  for  more  than 
three  regular  pentagons  to  meet  at  a  vertex  so  as  to  form 
a  polyhedral  ^  ?     Why  ? 

Are  any  other  regular  convex  polyhedrons  possible 
whose  faces  are  regular  pentagons  ? 

SUG.  19.  How  many  degrees  in  each  Z.  of  a  regular 
hexagon  ?  Is  it  possible  for  three  or  more  regular  hexa- 
gons to  meet  at  a  vertex  so  as  to  form  a  polyhedral  ^  ? 
Why  ?  Are  any  regular  convex  polyhedrons  possible 
whose  faces  are  regular  hexagons  ? 

SUG.  20.  Are  any  regular  convex  polyhedrons  possible 
whose  faces  are  regular  polygons  of  more  than  six  sides  ? 

SUG.  21.  Five  regular  convex  polyhedrons  have  been 
enumerated;  are  any  others  possible  ? 

Therefore 

463.  SCHOLIUM.  The  five  regular  polyhedrons  may 
be  constructed  of  cardboard,  as  follows:  Cut  the  material 
in  the  shape  of  the  following  patterns,  then  cut  it  half 
through  along  the  lines  separating  the  polygons.  Fold  it 
over  and  join  the  edges.  Paste  a  strip  of  paper  neatly 
over  the  joined  edges. 


TETRAHEDRON. 


HEXAHEDRON. 


OCTAHEDRON. 


POLYHEDRONS.  325 


AAAAA 

vvvvv 


DODECAHEDRON.  ICOSAHEDRON. 

Ex.  307.  To  construct  a  regular  tetrahedron. 

SUG.  Construct  an  equilateral  A  for  the  base.  At  the 
center  erect  a  _L  .  With  a  radius  equal  to  a  side  of  the 
A  and  with  a  center  at  one  vertex  of  the  A,  cut  off  the 
perpendicular. 

Ex.  308.  To  construct  a  regular  hexahedron. 

Ex.  309.  To  construct  a  regular  octahedron. 

SUG.  Construct  a  square,  and  at  its  center  erect  a  _L . 
With  a  vertex  of  the  square  as  a  center,  and  a  radius 
equal  to  one  side  of  the  square,  cut  off  the  _L  on  each 
side  of  the  square. 

Ex.  310.  To  construct  a  regular  icosahedron. 

SUG.  Construct  a  regular  pentagon,  and  at  its  center 
erect  a  _L .  With  a  vertex  of  the  pentagon  as  a  center, 
and  a  radius  equal  to  one  side  of  the  pentagon,  cut  off 
the  _L.  Join  the  point  on  the  _L  to  each  vertex  of  the 
pentagon  thus  forming  a  pentagonal  pyramid.  The  poly- 
hedral ^.  at  the  vertex  is  one  of  the  polyhedral  ^s  of 
the  icosahedron  required.  In  a  similar  manner,  construct 
polyhedral  ^s  at  the  vertices  of  the  pentagon  first  drawn. 


PROPOSITIONS  IN  CHAPTER  VII. 


PROPOSITION  I. 

The  lateral  edges  of  a  prism  are  equal  and  parallel,  and  make 
equal  angles  with  the  plane  of  either  base. 

PROPOSITION  II. 

Sections  of  a  prism  made  by  parallel  planes  are  polygons  which 
are  equal  in  all  respects. 

PROPOSITION  III. 

The  lateral  area  of  a  prism  equals  the  product  of  a  lateral  edge  by 
the  perimeter  of  a  right  section  of  the  prism. 

PROPOSITION  IV. 

Two  prisms  are  equal  in  all  respects  if  the  three  faces  about  a  tri- 
hedral angle  of  one  are  respectively  equal  in  all  respects  to  the  three 
faces  about  a  trihedral  angle  of  the  other,  and  these  faces  are  simi- 
larly placed  in  the  two  figures. 

PROPOSITION  V. 

Two  truncated  prisms  are  equal  in  all  respects  if  the  three  faces 
about  a  trihedral  angle  of  one  are  respectively  equal  in  all  respects  to 
the  three  faces  about  a  trihedral  angle  of  the  other,  and  these  faces 
are  similarly  placed  in  the  two  figures. 

PROPOSITION  VI. 

An  oblique  prism  is  equal  in  magnitude  to  a  right  prism  whose 
base  is  a  right  section  of  the  oblique  prism  and  whose  altitude  is  a 
lateral  edge  of  the  oblique  prism. 

PROPOSITION  VII. 

Opposite  faces  of  a  parallelepiped  are  parallelograms  which  are 
equal  in  all  respects. 


POLYHEDRONS.  327 


PROPOSITION  VIII. 

A  plane  passed  through  the  diagonally  opposite  edges  of  a  paral- 
lelopiped  divides  it  into  two  triangular  prisms  which  are  equal  in 
magnitude. 

PROPOSITION  IX. 

Any  parallelepiped  is  equal  in  volume  to  a  rectangular  parallele- 
piped having  an  equal  altitude  and  a  base  equal  in  area. 

PROPOSITION  X. 

Two  rectangular  parallelepipeds  having  equal  bases  have  the  same 
ratio  as  their  altitudes. 

PROPOSITION  XI. 

The  number  of  units  of  volume  in  a  rectangular  parallelepiped  is 
equal  to  the  product  of  the  number  of  linear  units  in  the  edges  meet- 
ing at  any  vertex. 

PROPOSITION  XII. 

The  volume  of  any  parallelepiped  is  equal  to  the  product  of  the 
area  of  its  base  by  its  altitude. 

PROPOSITION  XIII. 

The  volume  of  a  triangular  prism  is  equal  to  the  product  of  the 
area  of  its  base  by  its  altitude. 

PROPOSITION  XIV. 

The  volume  of  any  prism  is  equal  to  the  product  of  the  area  of  its 
base  by  its  altitude. 

PROPOSITION  XV. 

If  a  pyramid  be  cut  by  a  plane  parallel  to  the  base: 
I.     The  edges  and  altitude  are  divided  proportionally. 
II.     The  seetion  is  a  polygon  similar  to  the  base 

PROPOSITION  XVI. 

Two  triangular  pyramids  having  equal  altitudes  and  bases  equal  in 
area,  are  equal  in  volume. 


328  SOLID   GEOMETRY. 


PROPOSITION  XVII. 

The  volume  of  a  triangular  pyramid  is  one  third  the  volume  of  a 
triangular  prism  having  the  same  base  and  altitude. 

PROPOSITION  XVIII. 

The  volume  of  any  pyramid  is  equal  to  one  third  of  the  product  of 
the  area  of  its  base  by  its  altitude. 

PROPOSITION  XIX. 

The  volume  of  the  frustum  of  a  triangular  pyramid  is  equal  to  the 
sum  of  the  volumes  of  three  triangular  pyramids  whose  common  alti- 
tude is  the  altitude  of  the  frustum  and  whose  bases  are  respectively 
the  upper  base  of  the  frustum,  the  lower  base  of  the  frustum,  and  a 
mean  proportional  between  the  bases  of  the  frustum. 

PROPOSITION  XX. 

The  volume  of  the  frustum  of  any  pyramid  is  equal  to  the  sum  of 
the  volumes  of  three  pyramids  whose  common  altitude  is  the  altitude 
of  the  frustum  and  whose  bases  are  respectively  the  upper  base  of 
the  frustum,  the  lower  base  of  the  frustum,  and  a  mean  proportional 
between  the  bases  of  the  frustum. 

PROPOSITION  XXI. 

The  lateral  area  of  a  regular  pyramid  is  equal  to  one  half  the  prod- 
uct of  the  perimeter  of  the  base  by  the  slant  height. 

PROPOSITION  XXII. 

The  lateral  area  of  the  frustum  of  a  regular  pyramid  equals  the 
product  of  the  slant  height  by  one  half  the  sum  of  the  perimeters  of 
the  two  bases. 

PROPOSITION  XXIII. 
Only  five  regular  convex  polyhedrons  are  possible. 


CHAPTER  VIII. 
THE  THREE  ROUND  BODIES. 


THE  CYLINDER. 

464.  A  cylindrical  surface  is  a  surface  formed  by  a 
moving  straight  line  which  al- 
ways remains  parallel  to  itself 
and  continually  touches  a  given 
curved  line.  The  moving  straight 
line  is  called  the  generatrix,  and 
the  given  curved  line  is  called 
the  directrix. 


465.  A  straight  line  in  a  cylindrical  surface  which 
occupies  one  of  the  positions  of  the  generatrix  is  called 
an  element  of  the  surface. 

466.  A  cylinder  is  a  figure  bounded  by  a  cylindrical 
surface  and  two  parallel  planes.     The 

plane  surfaces  are  called  the  bases  of 
the  cylinder,  and  the  cylindrical  sur- 
face is  called  the  lateral  surface  of 
the  cylinder.  The  perpendicular  dis- 
tance between  the  bases  is  the  altitude 
of  the  cylinder. 

467.  A  right  section  of  a  cylinder  is  the  intersection 
of  the  cylinder  and  a  plane  perpendicular  to  an  element 
of  the  surface. 


330  SOLID    GEOMETRY. 

468.  A  circular  cylinder  is  a  cylinder  whose  bases 
are  circles. 

469.  A  right  cylinder  is  a  cylinder  whose  elements 
are  perpendicular  to  its  bases. 

470.  A  cylinder  of  revolution  is  a  right  circular 
cylinder.     It  may  be  generated  by  the  revolution  of  a 
rectangle  about  one  of  its  sides. 

471.  Similar  cylinders  of  revolution  are  cylinders 
generated  by  the  revolution  of  similar  rectangles  about 
homologous  sides. 

472.  A  tangent  line  to  a  cylinder  is   a  line  which 
touches  the  cylindrical  surface  in  a  point  but  does  not 
intersect  the  surface  at  that  point. 

473.  A  tangent  plane  to  a  cylinder  is  a  plane  which 
touches  the  cylindrical  surface  along  one  of  the  elements 
but  does  not  intersect  the  surface  along  that  element. 

PROPOSITION  I.    THEOREM. 

474.  Every  section  of  a  cylinder  made  by  a  plane 
containing  an  element  is  a  parallelogram. 


Let  the  plane  A  D  contain  the  element  A  B, 


THE  CYLINDER.  331 


To  prove  that  the  section  A  B  C  D  is  a  parallelogram. 

SUG.  1.  D  is  one  point  common  to  the  plane  and  the 
surface  of  the  cylinder.  If  a  line  be  drawn  through  D  || 
to  B  A  will  it  lie  in  the  plane  A  D  ?  Why  ?  Will  it 
lie  in  the  surface  of  the  cylinder  ?  Why  ?  Will  it  coin- 
cide with  the  intersection  D  C?  How,  then,  is  D  C  re- 
lated to  B  A  ? 
.  SUG.  2.  How  is  A  C  related  to  B  D  ?  Give  auth. 

SUG.  3.  What  kind  of  quadrilateral  is  A  B  D  C1 

Therefore 

475.  COROU,ARY.     Every  section  of  a  right  cylinder 
containing  an  element  is  a  rectangle. 


Ex.  311.  Find  the  lateral  area  of  a  regular  pentagonal 
pyramid  whose  basal  edge  is  2  ft.  and  whose  slant  height 
is  1  yd. 

Ex.  312.  Find  the  total  area  of  a  regular  quadrangular 
pyramid  whose  basal  edge  is  8  ft.  and  whose  lateral  edge 
is  5  ft. 

Ex.  313.  Find  the  volume  of  a  regular  quadrangular 
pyramid  whose  basal  edge  is  8  ft.  and  whose  slant  height 
is  5  ft. 

Ex.  314.  Find  the  volume  of  a  regular  triangular  pyr- 
amid whose  basal  edge  is  4  ft.  and  whose  altitude  is 
1/3  ft. 

Ex.  315.  A  pyramid  whose  altitude  is  ^16"  ft.  is  cut 
into  two  parts  of  equal  volume  by  a  plane  parallel  to  the 
base.  Find  the  distance  of  the  cutting  plane  from  the 
vertex. 


332  SOLID   GEOMETRY. 

PROPOSITION  II.    THEOREM. 
476.  The  bases  of  a  cylinder  are  equal. 


Let  AEbea  cylinder  ivhose  bases  are  ABC  and  I>EF. 

To  prove  that  the  bases  ABC  and  D  E  F  are  equal. 

SUG.  1.  I/et  B  and  Cbe  any  two  points  in  the  perime- 
ter of  the  upper  base,  and  let  B  F  be  a  plane  containing 
the  line  B  C  and  the  element  B  E.  Compare  B  C  and  E  F. 

SUG.  2.  Let  A  represent  any  other  point  in  the  perim- 
eter of  the  upper  base,  and  A  D  the  element  through  A. 
Pass  a  plane  through  the  ||  lines  A  D  and  B  E ;  also  a 
plane  through  the  ||  lines  A  D  and  C  F.  Compare  B  A 
and  E  D  ;  also  C  A  and  F  D. 

SUG.  3.  Compare  As  A  B  C  and  D  E  F.     Give  auth. 

SUG.  4.  Consider  the  upper  base  placed  upon  the  lower 
base  with  B  C  upon  E  F.  Where  will  the  point  A  fall  ? 
Since  A  is  any  point  in  the  perimeter  of  the  upper  base, 
where  will  trie  perimeter  of  the  upper  base  lie  with  re- 
spect to  the  perimeter  of  the  lower  base  ? 

Therefore 

477.  COROU,ARY  I.  Any  two  parallel  sections  cut- 
ting all  the  elements  of  a  cylinder  are  equal. 


THE    CYLINDER. 


333 


478.  COROLLARY  II.     All  sections  of  a  circular  cyl- 
inder- parallel  to  the  bases  are  equal  to 

the  bases,  and  the  straight  line  which 
connects  the  centers  of  the  bases  passes 
through  the  centers  of  the  sections. 

SUG.   Draw  two  diameters  of  one  of 
the  bases,  and  through  these  diameters 
and   elements    of    the    cylinder    pass 
planes.     These  planes  will  intersect  the  sections  in  diam- 
eters.    Why  ? 

479.  A  cylinder  is  inscribed  in  a  prism  when  each 
lateral  face  of  the  prism  is  tan- 
gent to  the  cylinder  and  the  bases 

of    the    prism   circumscribe  the 
bases  of  the  cylinder. 

When  a  cylinder  is  inscribed 
in  a  prism  the  prism  is  said  to  be 
circumscribed  about  the  cyl- 
inder. 

480.  A   cylinder  is  circumscribed  about  a  prism 
when  each  edge  of  the  prism  is  an 

element  of  the  cylinder,  and  the 
bases  of  the  prism  are  inscribed  in 
the  bases  of  the  cylinder. 

When  a  cylinder  is  circumscribed 
about  a  prism  the  prism  is  said  to 
be  inscribed  in  the  cylinder. 


Ex.  316.  The  base  of  a  pyramid  is  a  square  whose  side 
is  6  ft. ,  and  the  lateral  area  of  the  pyramid  is  f  of  its 
total  area;  find  the  altitude  and  the  slant  height  of  the 
pyramid. 


334:  SOLID  GEOMETRY. 

PROPOSITION  III.     THEOREM. 

481.  The  area  of  the  lateral  surface  of  a  cylinder  is 
equal  to  the  perimeter  of  a  right  section  multiplied  by 
an  element  of  the  surface. 

A 


Let  AGbea  cylinder,  KM  a  right  section,  and  AEan 
element. 

To  prove  that  the  area  of  the  lateral  surface  equals  the 
perimeter  of  the  section  KM  multiplied  by  A  B. 

SUG.  1.  Inscribe  a  prism  within  the  cylinder. 

SUG.  2.  What  is  the  lateral  area  of  the  prism  equal  to  ? 
(Art.  410.) 

SUG.  3.  Let  the  number  of  sides  of  the  bases  of  the 
prism  be  increased  by  bisecting  the  arcs  subtended  by  the 
sides  and  joining  the  points  of  bisection  to  the  adjacent 
vertices. 

SUG.  4.  As  the  number  of  faces  of  the  prism  increases 
the  lateral  area  varies.  Why  ? 

SUG.  5.  What  is  the  limit  of  the  lateral  area  of  the 
prism  ? 

SUG.  6.  The  perimeter  of  the  polygon  inscribed  in  the 
right  section  approaches  what  limit  ? 


THE    CYLINDER. 


335 


SUG.  7.  As  the  lateral  area  of  the  prism  increases  what 
does  it  remain  equal  to  ?  (Art.  410.) 

SUG.  8.  What,  then,  is  the  area  of  the  lateral  surface 
of  the  cylinder  equal  to  ? 

Therefore 

482.  COROLLARY  I.  The  lateral  area  of  a  cylinder  of 
revolution  equals  the  perimeter  of  the  base  multiplied  by 
the  altitude. 

This  fact  may  be  expressed  by  the  formula 


wherein  A  represents  the  area  of  the  lateral  surface,  H 
the  altitude,  and  R  the  radius  of  the  base  of  the  cylinder 
of  revolution. 

483.  COROLLARY  II.  The  lateral  areas  of  similar  cyl- 
inders of  revolution  are  to 
each  other  as  the  squares  of 
their  altitudes  or  as  the 
squares  of  the  radii  of  their 
bases. 

A,  H  and  R  repre~ 


sent  respectively  the  area, 

altitude,  and  radius  of  the 

base  of  one  cylinder  and  a,  h  and  r  the  area,  altitude,  and 

radius  of  the  base  of  the  other  cylinder. 

Then  A= 

and  a= 

A  C)    —^     J~)     T_T  JO          T  y 

S±   &  71-  J\.  JT2  J\,      ft 

a  ~~  %nrh        '  r    ~h~° 


Hence  —  = 


But 


•7-  —  X«   why  ? 


Hence  —  =  -^,  or  ^-. 

a  r2          h* 


336  SOLID    GEOMETRY. 

MODEL. 

PROPOSITION  III.    THEOREM. 

484.  The  area  of  the  lateral  surface  of  a  cylinder  is 
equal  to  the  perimeter  of  a  right  section  multiplied  by 
an  element  of  the  surface. 

B 


Let  A  G^e  a  cylinder,  KM  a  right  section,  and  AEan 
element. 

To  prove  that  the  area  of  the  lateral  surface  equals  the 
perimeter  of  the  section  K  M  multiplied  by  A  E. 

Inscribe  a  prism  within  the  cylinder. 

The  lateral  area  of  this  prism  is  equal  to  the  perimeter 
of  the  right  section  K  L  M  N  multiplied  by  an  edge  of 
the  prism,  as  A  E.  (Art.  410.) 

Now,  let  the  number  of  sides  of  the  bases  of  the  prism 
be  increased  by  bisecting  the  arcs  subtended  by  the  sides 
and  joining  the  points  of  bisection  to  the  adjacent  ver- 
tices. As  the  number  of  sides  of  the  bases  of  the  prism 
increases  the  number  of  faces  of  the  prism  increases. 

As  the  number  of  faces  of  the  prism  increases  the  lat- 
eral area  of  the  prism  also  increases;  for  the  perimeter  of 
a  right  section  increases,  and  hence  this  perimeter  multi- 
plied by  an  edge  of  the  prism  increases. 


THE    CYLINDER.  837 


The  lateral  area  of  the  prism  approaches  the  lateral 
area  of  the  cylinder  as  a  limit. 

The  perimeter  of  a  right  section  of  the  prism  ap- 
proaches the  perimeter  of  a  right  section  of  the  cylinder 
as  a  limit. 

Hence,  the  lateral  area  of  a  prism  is  one  variable,  and 
the  perimeter  of  a  right  section  of  the  prism  multiplied 
by  an  edge  is  another  variable,  and  these  two  variables 
are  always  equal.  (Art.  410.) 

Hence  the  limits  of  these  variables  are  also  equal. 
(Art.  170.) 

Therefore  the  area  of  the  lateral  surface  of  a  cylinder 
is  equal  to  the  perimeter  of  a  right  section  multiplied  by 
an  element  of  the  surface. 


Ex.  317.  If  from  any  point  within  an  equilateral  tri- 
angle perpendiculars  be  drawn  to  the  three  sides,  the  sum 
of  those  perpendiculars  is  equal  to  the  altitude  of  the  tri- 
angle. 

SUG.  Divide  the  given  A  into  three  As  whose  common 
vertex  is  at  the  point  from  which  the  _Ls  are  drawn. 

Ex.  318.  If  from  any  point  within  a  regular  tetrahe- 
dron perpendiculars  be  drawn  to  the  four  faces,  the  sum 
of  those  perpendiculars  is  equal  to  the  altitude  of  the 
tetrahedro  i. 

SUG.  Divide  the  tetrahedron  into  four  triangular  pyra- 
mids whose  common  vertex  is  at  the  point  from  which 
the  _Ls  are  drawn. 

Ex.  319.  The  lateral  areas  of  similar  pyramids  have 
the  same  ratio  as  the  squares  of  their  altitudes,  or  as  the 

squares  of  any  two  homologous  edges. 
22— Geo. 


338  SOLID   GEOMETRY. 


PROPOSITION  IV.    THEOREM. 

485,   The  volume  of  a  cylinder  is  equal  to  the  area 
of  its  base  multiplied  by  its  altitude. 


Let  A  G  be  a  cylinder  whose  base  is  EG. 

To  prove  that  the  volume  of  the  cylinder  A  G  equals  the 
area  of  the  base  E  G  multiplied  by  the  altitude. 

SUG.  1.  Inscribe  a  prism  within  the  cylinder  and  then 
let  the  number  of  the  faces  of  the  prism  increase  indef- 
initely. What  is  the  volume  of  the  prism  always  equal 
to?  (Art.  436.) 

SUG.  2.  What  two  variables  are  always  equal  ? 

SUG.  3.  What  is  the  limit  of  the  volume  of  the  prism  ? 

SUG.  4.  What  is  the  limit  of  the  area  of  the  base  of 
the  prism  multiplied  by  its  altitude? 

Therefore  - 

486.  COROLLARY  I.  The  volume  of  a  cylinder  of  rev- 
volution  may  be  expressed  by  the  formula 


wherein  V  represents  the  volume,  //'the  altitude,  and  R 
the  radius  of  the  base  of  the  cylinder  of  revolution. 


CONES.  339 


486.  COROLLARY  II.     The  volumes  of  similar  cylin- 
ders of  revolution  are  to  each  other  as  the  cubes  of  their 
altitudes  or  the  cubes  of  the  radii  of  their  bases.     Let 
F,  H  and  R  represent  respectively  the  volume,  altitude 
and  radius  of  the  base  of  one  cylinder,  and  v,  h  and  r  the 
volume,  altitude  and  radius  of  the  base  of  the  other  cyl- 
inder. 

Then 
and 

Hence  • — 

But 

Hence  - —  =  — ,  or  -T^-. 

h* 

CONES. 

487.  A  conical  surface  is  a  curved  surface  formed  by 
a    moving    straight    line   which   passes 

through  a  given  fixed  point  and  contin- 
ually touches  a  given  curve.  The  mov- 
ing straight  line  is  called  the  generatrix, 
and  the  given  curve  is  called  the  di- 
rectrix. 

488.  A  straight  line  in  a  conical  sur- 
face which  occupies  one  of  the  positions  of  the  generatrix 
is  called  an  element  of  the  surface. 

489.  A  cone  is  a  figure  bounded  by 
a  conical  surface  and  a  plane.     The 
plane  is  called  the  base  of  the  cone, 
the  conical  surface  is  called  the  lateral 
surface  or  the  convex  surface  of  the 
cone,  and  the  point  in  which  the  ele- 
ments meet  is  called  the  vertex  of  the 


34:0 


SOLID    GEOMETRY. 


cone.     The   perpendicular  distance  from  the  vertex   to 
the  base  is  the  altitude  of  the  cone. 

490.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 
A  straight  line  joining  the  vertex  with  the  center  of  the 
base  is  called  the  axis  of  the  cone. 

491.  A  right  circular  cone  is  a  cone  whose  axis  is 
perpendicular  to  its  base.     A  right  cir- 
cular cone  is  a  cone  of  revolution.     It 

may  be  generated  by  revolving  a  right 
triangle  about  one  of  its  legs.  The 
hypotenuse  in  any  one  of  its  positions  is 
an  element  of  the  cone  and  is  called  the 
slant  height  of  the  cone. 

492.  Similar  cones  of  revolution  are  cones  that  may 
be  generated  by  the  revolution  of  similar  right  triangles 
about  homologous  sides. 

493.  A  tangent  line  to  a  cone  is  a  line  which  touches 
the  conical  surface  in  a  point  but  does  not  intersect  the 
surface  at  that  point. 

494.  A  tangent  plane  to  a  cone  is  a  plane  which 
touches  the  conical  surface  along  one  of  the  elements  but 
does  not  intersect  the  surface  along  that  element. 

495.  A  frustum  of  a  cone  is 

that  part  of  a  cone  included  be- 
tween the  base  and  a  section  paral- 
lel to  the  base.  The  base  of  the 
cone  is  the  lower  base  of  the  frus. 
turn,  the  section  parallel  to  the  base 
of  the  cone  is  the  upper  base  of 


CONES. 


341 


the  frustum.  The  perpendicular  distance  between  the 
bases  is  the  altitude  of  the  frustum.  The  slant  height 
of  the  frustum  of  a  cone  of  revolution  is  that  part  of  the 
slant  height  of  the  cone  which  is  included  between  the 
bases  of  the  frustum. 

PROPOSITION  V.    THEOREM. 

496.  Every  section  of  a  cone  made  by  a  plane  pass- 
ing through  the  vertex  is  a  triangle. 


Let  AS C  be  a  section  of  a  cone  by  a  plane  passing 
through  the  vertex  A. 

To  prove  that  A  B  C  is  a  triangle. 

SuG.  1.  What  kind  of  a  line  is  A  C?     Why?     (Art. 
488.)     What  kind  of  a  line  is  A  B  ?  B  C?     Why  ? 
SUG.  2.  What  is  the  figure  A  B  C? 
Therefore 


Hx.  320.  The  volumes  of  two  similar  pyramids  have 
the  same  ratio  as  the  cubes  of  their  altitudes,  or  as  the 
cubes  of  any  two  homologous  edges. 


342  SOLID   GEOMETRY. 

PROPOSITION  VI.    THKORKM. 

497,  Every  section  of  a  circular  cone  made  by  a 
plane  parallel  to  the  base  is  a  circle. 


Let  BOCbe  the  base  of  a  circular  cone,  and  ESFa  sec- 
tion parallel  to  the  base. 

To  prove  that  the  section  E  S  Fis  a  circle. 

SUG.  1.  ]>t  A  O  be  the  axis  of  the  cone,  and  5  the 
point  in  which  the  axis  intersects  the  section  E  S  F. 
I^et  E  and  F  be  any  two  points  whatever  in  the  perimeter 
of  the  section  E  S  F. 

SUG.  2.  Through  the  axis  A  O  and  the  point  E  pass 
a  plane  intersecting  the  base  in  the  line  O  B  and  the  sec- 
tion in  the  line  S  E.  Also  through  the  axis  A  O  and 
the  point  F  pass  a  plane  intersecting  the  base  in  the  line 
O  C,  ancl  the  section  in  the  line  6*  F. 

SUG.  3.  5  E  is  ||  to  O  B,  and  5  F  is  ||  to  O  C.  Give 
auth.  (Art.  361.) 

AS  S  E 

SUG.  4.  Compare  —r-^  with  -=—=.       Give  auth.     Also 

Si  U  U  £> 

AS       ^SF 
compare  _  with  ^ 


CONES.  343 


9*  f?  *?  /** 

SUG.  5.  Compare     -    with  -.     Give  auth. 


SUG.  6.  The  base  of  the  cone  is  a  circle  whose  center 
is  at  O.  Why?  (Art.  490.) 

SUG.  7.  Since  O  B  and  O  C  are  equal,  S  E  and  5  .F 
are  how  related  ? 

SUG.  8.  Since  E  and  ^  are  any  two  points  in  the  per- 
imeter of  the  section,  what  kind  of  a  figure  is  the  sec- 
tion E  S  F? 

Therefore  -- 

498.  COROLLARY.  *The  axis  of  a  circular  cone  passes 
through  the  centers  of  all  sections  parallel  to  the  base. 

499.  A  cone  is  inscribed  in  a  pyr- 
amid when  the  vertex  of  the  cone  co- 
incides with  the  vertex  of  the  pyramid 
and  the  base  of  the  cone  is  inscribed 
in  the  base  of  the  pyramid. 

When  a  cone  is  inscribed  in  a  pyra- 
mid the  pyramid  is  said  to  be  circumscribed  about  the 
cone. 

500.  A    cone    is    circumscribed 
about  a  pyramid  when  the  vertex  of 
the  cone  coincides  with  the  vertex  of 
the  pyramid  and  the  base  of  the  cone  is 
circumscribed  about  the  base  of  the 
pyramid. 

When  a  cone  is  circumscribed  about  a  pyramid  the 
pyramid  is  *aid  to  be  inscribed  in  the  cone. 


Ex.  321.  If  a  cone  is  inscribed  in  a  pyramid  prove  that 
each  lateral  face  of  the  pyramid  is  tangent  to  the  cone. 


344  SOLID   GEOMETRY. 

PROPOSITION  VII.    THEOREM. 

501.  The  area  of  the  convex  surface  of  a  cone  of 
revolution  is  equal  to  one  half  the  product  of  the  per- 
imeter of  its  base  by  its  slant  height. 


Let  A-B  CDEbea  cone  of  revolution. 

To  prove  that  the  area  of  the  convex  surface  of  the  cone  is 
equal  to  one  half  the  product  of  the  perimeter  of  its  base  by 
its  slant  height. 

SUG.  1.  Inscribe  a  regular  pyramid  within  the  cone. 

SUG.  2.  What  is  the  lateral  area  of  the  pyramid  equal 
to? 

Complete  the  demonstration  by  a  method  similar  to 
that  of  Prop.  III. 

Therefore  - 

502.  COROLLARY  I.  The  truth  of  the  theorem  may 
be  expressed  by  the  formula 


in  which  A  represents  the  area  of  the  convex  surface,  R 
the  radius  of  the  base,  and  6"  the  slant  height  of  the  cone 
of  revolution. 


CONES. 


345 


503.  COROLLARY  II.  The  lateral  areas  of  two  similar 
cones  of  revolution  are  to  each  other  as  the  squares  of 
their  slant  heights,  or  the  squares  of  their  altitudes,  or 
the  squares  of  the  radii  of  their  bases. 

See  method  of  Art.  483. 


Ex.  322.  Prove  the  above  proposition  by  using  a  cir- 
cumscribed instead  of  an  inscribed  pyramid. 

Ex.  323.  If  a  cone  is  circumscribed  about  a  pyramid 
prove  that  each  lateral  edge  of  the  pyramid  is  an  element 
of  the  cone. 

PROPOSITION  VIII.    THEOREM. 

504.  The  area  of  the  convex  surface  of  a  frustum 
of  a  cone  of  revolution  is  equal  to  one  half  the  product 
of  its  slant  height  by  the  sum  of  the  perimeters  of  its 
"bases. 


SUG.  Express  the  lateral  area  of  the  circumscribed 
frustum  of  a  pyramid.  Let  the  number  of  lateral  faces 
of  the  frustum  of  the  pyramid  increase,  and  employ  the 
method  of  limits  in  a  manner  similar  to  that  of  Prop.  III. 

Therefore 


34:6  SOLID   GEOMETRY. 

PROPOSITION  IX.     THEOREM. 

505.  The  volume  of  a  cone  equals  one  third  of  the 
product  of  the  area  of  its  base  by  its  altitude. 


Let  A-B  CDE  represent  a  cone. 

To  prove  that  the  volume  of  the  cone  equals  one  third  the 
product  of  the  area  of  its  base  by  its  altitude. 

SuG.  1.  Inscribe  a  pyramid  within  the  cone. 
SuG.  2.  What  is  the  volume  of  the  pyramid  equal  to  ? 
SUG.  3.  Complete  the  demonstration  by  a  method  sim- 
ilar to  that  of  Prop.  IV. 
Therefore  - 

506.  COROI^ARY  I.     The  volume  of  a  circular  cone 
may  be  expressed  by  the  formula 


wherein  V  represents  the  volume,  H  the  altitude,  and  R 
the  radius  of  the  base  of  the  cone. 

507.  COROU,ARY  II.  The  volumes  of  similar  cones 
of  revolution  are  to  each  other  as  the  cubes  of  the  radii 
of  their  bases,  or  the  cubes  of  the  altitudes,  or  the  cubes 
of  the  slant  heights. 

SUG.  Prove  this  by  a  method  similar  to  that  used  in 
Art.  486. 


THE    SPHERE.  347 


THE  SPHERE. 

508.  A  sphere  is  a  solid  bounded  by  a  surface  all 
the  points  of  which  are  equally  distant  from  a  fixed  point 
within  it. 

The  fixed  point  is  called  the  center  of  the  sphere,  and 
the  bounding  surface  is  called  the  surface  of  the  sphere. 

A  sphere  may  be  generated  by  the  revolution  of  a  semi- 
circle about  its  diameter. 

In  higher  branches  of  mathematics  the  word  sphere  is  used  to 
denote  what  is  here  spoken  of  as  the  surface  of  the  sphere,  but  in  this 
book  a  sphere  is  considered  as  a  solid  instead  of  the  surface  which 
bounds  the  solid. 

509.  The  radius  of  a  sphere  is  any  straight  line  drawn 
from  the  center  to  the  surface  of  the  sphere. 

The  diameter  of  a  sphere  is  any  straight  line  drawn 
through  the  center,  terminated  both  ways  by  the  surface 
of  the  sphere. 

From  the  definition  of  a  sphere,  all  radii  of  the  same 
sphere  are  equal,  and  since  a  diameter  is  twice  the  radius, 
it  follows  that  all  diameters  of  the  same  sphere  are  equal. 

510.  A  line  is  tangent  to  a  sphere  when  it  has  only 
one  point  in  common  with  the  sphere. 

A  plane  is  tangent  to  a  sphere  when  it  has  only  one 
point  in  common  with  the  sphere. 

Two  spheres  are  tangent  to  each  other  when  they 
have  only  one  point  in  common. 


Ex.  324.  The  slant  height  of  a  cone  of  revolution  is 
13  ft.,  and  its  altitude  is  12  ft.  Find  the  lateral  surface 
and  the  volume  of  the  cone. 


348  SOLID   GEOMETRY. 

PROPOSITION  X.     THEOREM. 

511.  Every  section  of  a  sphere  made  by  a  plane  is  a 
circle. 


Let  the  plane  MN  intersect  the  sphere  O,  in  the  section 
A  SB. 

To  prove  that  the  section  A  S  B  is  a  circle. 

SUG.  1.  From  the  center  of  the'sphere  O  drop  a  J_  O  S 
to  the  section,  and  from  5  draw  SA  and  SB  to  any  two 
points  in  the  perimeter  of  the  section.  Connect  A  and  O  ; 
also  B  and  O. 

SUG.  2.  Compare  A  O  and  O  B.  Compare  As  A  O  S 
and  B  O  S. 

SUG.  3.  Compare  5  A  and  5  B. 

SUG.  4.  What  is  the  section  ? 

Therefore 

QUERY.  Suppose  a  plane  passes  through  the  center  of 
a  sphere,  is  the  section  a  circle  ?  Why  ?  The  section  of 
a  sphere  made  by  a  plane  is  called  a  circle  of  the  sphere. 

512.  COROU,ARY  I.  The  line  drawn  from  the  center 
of  a  sphere  to  the  center  of  a  circle  of  a  sphere  is  perpen- 
dicular to  the  circle  of  the  sphere. 


THE    SPHERE.  34:$ 


513.  COROLLARY  II.     If  two  circles  of  a  sphere  are 
equally  distant  from  the  center,  they  are  equal. 

514.  COROLLARY  III.     Of  two  circles  of  a  sphere  un- 
equally distant  from  the  center,  that  one  is  greater  which 
is  nearer  the  center. 

515.  COROLLARY  IV.     All  circles  of  a  sphere  which 
contain  the  center  are  equal. 

516.  A  great  circle  of  a  sphere  is  one  which  contains 
the  center  of  the  sphere. 

A  small  circle  of  a  sphere  is  one  which  does  not  con- 
tain the  center  of  the  sphere. 

517.  The  poles  of  a  circle  of  a  sphere  are  the  extrem- 
ities of  the  diameter  which  is  perpendicular  to  the  plane 
of  the  circle.     This  diameter  is  often  called  the  axis  of 
the  circle. 


Ex.  325.  Prove  Prop.  VIII  by  considering  the  convex 
surface  of  the  frustum  of  a  cone  as  the  difference  between 
the  convex  surfaces  of  two  cones. 

SUG.  Let  A  denote  the  area  of  the  convex  surface,  R 
the  radius  of  the  base,  and  6"  the  slant  height  of  the 
larger  cone,  and  a,  r  and  s  the  area  of  the  convex  sur- 
face, the  radius  of  the  base,  and  the  slant  height  of  the 
smaller  cone. 

Then  A  —  itRS,  and  a  =  7trs. 

Find  A  —  a  to  agree  with  the  theorem. 

Ex.  326.  Prove  that  the  convex  surface  of  the  frustum 
of  a  cone  of  revolution  is  equal  to  the  circumference  of  a 
section  equidistant  from  the  bases  multiplied  by  the  slant 
height. 


350  SOLID    GEOMETRY. 

PROPOSITION  XI.    THEOREM. 

518.   Two  intersecting  great  circles  of  a  sphere  bi- 
sect each  other. 

B 


Let  A&CE  and  B  D  FE  be  two  great  circles  of  a 
sphere  intersecting  in  the  line  J>  E. 

To  prove  that  the  circles  A  DCE  and  B  DFE  bisect 
each  other. 

SUG.  1.  Is  the  center  of  the  sphere  in  the  plane  of  the 
QADCEl  Why? 

SUG.  2.  Are  all  points  in  the  circumference  of  the 
O  A  D  CE  equally  distant  from  the  center  of  the  sphere  ? 
Why? 

SUG.  3.  Locate  the  center  of  the  O  A  D  CE  with  re- 
spect to  the  center  of  the  sphere. 

SUG.  4.  Locate  the  center  of  the  O  B  DFE  with  re- 
spect to  the  center  of  the  sphere. 

SUG.  5.  Are  the  centers  of  the  Os^Z>C^  and  ^£>/^E 
jn  the  line  of  intersection  D  E? 

SUG.  6.  What  is  a  line  through  the  center  of  a  O 
called? 

SUG.  7.  Does  the  line  DE>  bisect  each  of  the  QsADCE 
DFE? 

Therefore 


THE    SPHERE.  351 


PROPOSITION  XII.     THEOREM. 

519.  Three  points  on  the  surface  of  a  sphere  deter- 
mine a  circle  of  the  sphere. 

SUG.  1.  How  many  points  determine  a  plane? 
SUG.  2.  A  plane  intersects  a  sphere  in  what  kind  of  a 
figure  ? 

Therefore 

520.  COROLLARY.     Two  points  on  the  surface  of  a 
sphere  determine  a  great  circle  of  a  sphere  if  the  points 
are  not  the  extremities  of  a  diameter  of  the  sphere. 

SUG.  What  third  point  is  known  in  addition  to  the 
two  given  points  ? 


Ex.  327.  Prove  Prop.  IX  by  circumscribing  a  pyramid 
about  the  cone. 

Ex.  328.  The  volume  of  the  frustum  of  any  cone  is 
equal  to  one  third  the  altitude  multiplied  by  the  sum  of 
the  upper  base,  the  lower  base,  and  a  mean  proportional 
between  the  two  bases.  (See  Art.  457.) 

Ex.  329.  The  total  area  of  a  right  circular  cylinder  is 
80  n  square  ft.  and  the  radius  of  the  base  is  5  ft.  Find 
the  altitude  of  the  cylinder. 

Ex.  330.  The  circumference  of  the  base  of  a  cone  of 
revolution  is  6  ft. ,  and  the  altitude  of  the  cone  is  n  ft. 
Find  the  volume  of  the  cone. 

'Ex.  331.  The  altitude  of  the  frustum  of  a  cone  is  \  the 
altitude  of  the  entire  cone.  Find  the  relation  between 
the  volumes  of  the  frustum  and  the  entire  cone. 

Ex.  332.  Two  circular  cylinders  have  the  same  altitude 
but  the  volume  of  one  is  four  times  the  volume  of  the 
other.  Find  the  relation  between  the  radii  of  the  bases 
of  the  two  cylinders. 


352  SOLID  GEOMETRY. 

PROPOSITION  XIII.     THEOREM. 

521.  The  shortest  distance  on  the  surface  of  a  sphere 
between  any  two  points  on  that  surface  is  the  arc,  not 
greater  than  a  semi-circumference,  of  a  great  circle 
which  joins  them. 


Let  A  and  B  be  two  points  on  the  surface  of  a  sphere, 
and  let  A  B  be  the  arc,  not  greater  than  a  semi-circle, 
which  joins  A  and  B,  and  let  A  CMDB  be  any  other  line 
on  the  surface  of  the  sphere  joining  A  and  B. 

To  prove  that  A  B  is  less  than  A  CMDB. 

SuG.  1.  Connect  any  point  of  A  CMDB,  as  M,  with 
A  and  B,  by  arcs  of  great  Os,  as  MA  and-  MB.  Con- 
nect A,  Maud  B  with  O,  the  center  of  the  sphere. 

SuG.  2.  In  the  trihedral  Z.  O-A  MB  compare  the  face 
Z  A  0j5  with  the  sum  of  the  face  Zs  A  OM&n&MOB. 

SUG.  3.  Compare  the  arc  A  B  with  the  sum  of  the  arcs 
A  M  and  MB.  (Art.  174.) 

SUG.  4.  In  the  same  way  connect  C,  any  point  in  the 
line  ACM,  with  A  and  M  by  arcs  of  great  Os.  Also 
connect  D  in  the  line  MD  B  with  M  and  B  by  arcs  of 
great  Os. 

SUG.  5.  Compare  the  sum  of  the  great  O  arcs  A  C,  CM, 
M  D  and  D  B  with  the  sum  of  the  great  O  arcs  A  M  and 
MB,  and  also  with  A  B. 


THE    SPHERE.  353 


SUG.  6.  Continue  to  take  points  in  the  line  A  C M B D 
and  proceed  as  before.  The  sum  of  the  arcs  of  great  Os  is 
a  variable.  Why?  What  is  its  limit?  Why? 

SUG.  7.  How  does  each  succeeding  value  of  the  vari- 
able compare  with  the  one  preceding  it  ? 

SUG.  8.  Then,  how  must  A  C  M  D  B  compare  with 
A  SI 

Therefore  

522.  By  the  distance  between  two  points  on  the 
surface  of  a  sphere  is  meant  the  distance  measured  on 
the  arc  of  a  great  circle. 


Ex.  333.  The  portion  of  a  tetrahedron  cut  off  by  a 
plane  parallel  to  any  face  is  a  tetrahedron  similar  to  the 
given  tetrahedron. 

Ex.  334.  Find  the  volume  of  the  solid  generated  by 
revolving  an  equilateral  triangle  whose  side  is  6  ft.  about 
one  of  its  sides. 

SUG.  Notice  that  two  cones  are  generated  whose  bases 
coincide,  and  whose  vertices  are  on  opposite  sides  of  the 
common  base. 

Ex.  335.   Compare  the  volumes  of  the  solids  generated* 
by  revolving  a  rectangle  whose  dimensions  are  a  and  b 
successively  about  the  adjacent  sides. 

Ex.  336.  The  volume  of  a  cylinder  of  revolution  is 
equal  to  one  half  of  the  product  of  its  lateral  area  by  its 
radius. 

Ex.  337.  In  each  of  two  right  circular  cylinders  the 
altitude  is  equal  to  the  diameter,  and  the  volume  of  one 
is  YV  that  of  the  other.  Find  the  relation  of  their  alti- 
tudes. 

23— Geo. 


354 


SOLID    GEOMETRY. 


PROPOSITION  XIV.    THEOREM. 

523.  All  points  in  the  circumference  of  a  circle  are 
equally  distant  from  either  of  its  poles. 


Let  A  DC  B  be  the  circumference  of  a  circle,  and  E  and 
G  its  poles. 

To  prove  that  all  points  in  the  circumference  A  D  C  B 
are  equally  distant  from  E,  and  also  equally  distant  from  G. 

SUG.  1.  ]>t  M  be  the  center  of  the  O  A  D  CB.  The 
straight  line  passing  through  O  and  M  will  determine 
E  and  G,  the  poles  of  iheQADCfi.  Why  ?  (Arts. 
512  and  517.) 

SUG.  2.  The  distances  of  all  points  in  the  circumfer- 
ence A  D  CB  from  any  point  in  E  G  are  how  related  ? 
(Art.  345.) 

SUG.  3.  Then  the  distances  of  all  points  in  the  circum- 
ference A  D  C  B  from  E  are  how  related  ?  The  distances 
from  G  are  how  related  ? 

SUG.  4.  But  the  distances  considered  in  Sug.  3  are 
chords  of  great  Os.  Why  ? 

SUG.  5.  How,  then,  must  the  distances  measured  on 
the  surface  of  the  sphere  compare  ?  Give  auth. 

Therefore 


THE    SPHERE.  355 


524.  The  distance  of  any  point  in  the  circumference 
of  a  circle  from  the  nearer  pole  of  that  circle  is  called  the 
polar  distance  of  the  circle. 

525.  COROLLARY  I.     The  polar  distance  of  a  great 
circle  is  a  quadrant,  i.  e.t  an  arc  of  ninety  degrees. 

SUG.  What  is  the  Z.  at  the  center  of  a  sphere  that 
subtends  the  polar  distance  of  a  great  circle  ? 

526.  COROLLARY  II.     A  point  which  is  at  the  dis- 
tance of  a  quadrant  from  each  of  two  points  on  the  sur- 
face of  a  sphere  is  a  pole  of  a  great  circle  passing  through 
those  points. 

SUG.  L<et  E  be  the  distance  of  a  quadrant  from  both  R 
and  S.  Then  Z  R  OE  equals  what?  What  relation 
does  E  G  sustain  to  the  circle  R  O  S  ? 

527.  SCHOLIUM.     By  means  of  poles,  arcs  of  circles 
may  be  drawn  on  the  surface  of  a  sphere  in  much  the 
same  way  as  upon  a  plane  surface.     By  revolving  the 
arc  E  A  about  E  as  a  pole,  the  point  A  describes  the  cir- 
cumference ADCB.     By  revolving  the  quadrant  ER 
about  E  as  a  pole,  or  the  quadrant  G  R  about  G  as  a 
pole,  the  point  R  describes  the  circumference  of  the  great 
circle  R  S. 

To  describe  a  great  circle  arc  on  a  sphere  by  means  of 
dividers,  it  is  necessary  to  open  the  dividers  so  that  their 
extremities  touch  points  on  the  sphere  which  are  just 
ninety  degrees  apart,  then  with  the  extremity  of  one  leg 
of  the  dividers  fixed  at  some  point  on  the  sphere,  the 
extremity  of  the  other  leg  describes  a  great  circle  arc. 


350 


SOLID    GEOMETRY. 


PROPOSITION  XV.     PROBLEM. 
528.  Given  a  material  sphere,  to  find  its  radius. 


SuG.  1 .  Take  any  point  M  on  the  surface  of  the  sphere 
as  a  pole,  and  describe  a  circumference  A  B  S  C  on  the 
surface. 

SuG.  2.  Take  any  three  points  A,  B  and  Con  this  cir- 
cumference, and  by  means  of  dividers  construct  a  A  D  E  F 
equal  in  all  respects  to  the  A  A  B  C. 

SUG.  3.  Circumscribe  a  O  about  the  A  DBF,  and 
let  G  be  the  center  of  this  O. 

SUG.  4.  Draw  5  T  equal  to  the  radius  G  E,  and  at  5 
draw  an  indefinite  line  N  R  _L  to  T  S. 

SUG.  5.  With  T  as  a  center,  and  with  a  radius  equal 
to  the  straight  line  M  C,  describe  an  arc  cutting  the  line 
NRa.iR.  Join  T  and  R. 

SUG.  6.  At  T  erect  a  _L  to  T  R,  and  extend  this  J_ 
until  it  meets  N  R  at  N. 

SUG.  7.  Prove  that  N  R  is  equal  to  the  diameter  of 
the  sphere.  Find  the  radius  of  the  sphere. 


THE   SPHERE.  357 


PROPOSITION  XVI.     THEOREM. 

529.  A  plane  perpendicular  to  a  radius  of  a  sphere 
at  its  extremity  is  tangent  to  the  sphere. 


Let  A  J5  represent  a  plane  perpendicular  to  the  radius 
OMatM. 

To  prove  that  the  plane  A  B  is  tangent  to  the  sphere  O. 

SUG.  1.  Draw  any  other  line,  as  OP,  from  O  to  the  plane 
A  B,  and  connect  P  and  M. 

SUG.  2.   O  M  is  _L  to  P  M.     Why  ? 

SUG.  3.  Compare  O  P  and  O  M  in  respect  to  length. 
Give  auth. 

SUG.  4.  Where  then  must  the  point  P  lie  with  respect 
to  the  sphere  ? 

SUG.  5.  What  relation  does  the  plane  A  B  sustain  to 
the  sphere  O  ? 

Therefore 

530.  COROLLARY  I.  Any  straight  line  in  a  tangent 
plane  through  the  point  of  contact  is  tangent  to  the 
sphere. 


358 


SOLID   GEOMETRY. 


531.  COROU,ARY  II.     Any  line  perpendicular  to  the 
radius  of  a  sphere  at  its  extremity  is  tangent  to  the 
sphere. 

532.  The  angle  between  two  curves  which  pass 
through  the   same   point  is  the  angle  formed  by  two 
straight  lines  tangent  to  the  curves  at  that  point. 

533.  A  spherical  angle  is  an  angle  formed  by  the 
arcs  of  two  great  circles. 

PROPOSITION  XVII.    THEOREM. 

534.  A  spherical  angle  is  equal  to  the  dihedral 
angle  formed  by  the  planes  of  the  two  arcs,  and  is 
measured  by  the  arc  of  a  great  circle  described  from 
the  intersection  of  the  arcs  as  a  pole  and  intercepted 
between  them. 


Let  M.  S  and  N  S  represent  two  arcs  of  great  circles  in- 
tersecting at  S,  A  B  and  C  D  the  straight  lines  tangent  to 
the  arcs  at  S,  and  MNthe  arc  of  a  great  circle  whose  pole 
is  at  S  and  which  is  intercepted  between  the  arcs  S  N  and 
SM. 

To  prove  that  the  spherical  angle  MSN  equals  the  dihe- 
dral angle  MSPJV,  and  is  measured  by  the  arc  M  N. 


THE    SPHERE.  359 


SUG.  1.  Let  ON  be  the  radius  of  the  great  O  NS  T 
drawn  to  the  point  N,  and  O  M  the  radius  of  the  great 
O  M  S  R  drawn  to  the  point  M. 

SUG.  2.  ON  and  SD  are  each  in  the  plane  NS  T. 
Why? 

SUG.  3.  SD  is  how  related  to  5(9?  Why?  ON  is 
how  related  to  S  O  ?  Why  ?  5  D  and  O  N  are  how  re- 
lated to  each  other  ? 

SUG.  4.  6*  B  and  O  M  are  how  related  to  each  other  ? 
Why? 

SUG.  5.  Compare  Zs  BSD  and  M  O  N.     (Art.  362.) 

SUG.  6.  The  Z  MO  N  is  the  measure  of  the  dihedral 
ZMSPN.  Why? 

SUG.  7.  What  arc  is  the  measure  of  Z  MO  N?  Why  ? 

Therefore 

535.  A  spherical  polygon  is  a  portion  of  the  surface 
of  a  sphere  bounded  by  arcs  of  great        A 

circles;  as  A  B  CD. 

The  bounding  arcs,  A  B,  B  C,  etc., 
are  the  sides  of  the  polygon. 

As  the  sides  of  a  spherical  polygon 
are  arcs  of  great  circles  they  are  ex- 
pressed in  degrees. 

The  planes  of  the  sides  of  a  spherical  polygon  form  a 
polyhedral  angle  whose  vertex  is  at  the  center  of  the 
sphere.  Thus,  O-ABCD  is  a  polyhedral  angle  whose 
vertex  is  at  (9,  the  center  of  the  sphere. 

The  diagonal  of  a  spherical  polygon  is  the  arc  of  a 
great  circle  joining  any  two  vertices  not  adjacent. 

536.  A  spherical  triangle  is  a  spherical  polygon  of 
three  sides. 


3 CO  SOLID    GEOMETRY. 

537.  Spherical  triangles  are  right  angled,   isosceles, 
equilateral,  etc.,  in  the  same  cases  as  plane  triangles. 

538.  Arcs  of  great  circles  on  the  same  sphere  may  be 
superposed  in  the  same  way  as  straight  lines  are  super- 
posed. 

539.  Equal  spherical  angles  may  be  applied  to  each 
other  and  made  to  coincide  in  the  same  manner  as  plane 
angles. 

540.  Equal   spherical   triangles   are   spherical    tri- 
angles which,  by  placing  one  upon  the  other,  may  be 
made  to  coincide  in  every  part.     In  such  triangles  each 
part  of  one  is  equal  to  the  corresponding  part  of  the 
other. 

541.  Symmetrical  spherical  triangles  are  spherical 
triangles  in  which  the  parts  of  one 

are  equal  respectively  to  the  parts 
of  the  other,  but  arranged  in  re- 
verse order.  Such  triangles  can- 
not, except  in  the  case  stated  in 
Prop.  XVIII,  be  placed  one  upon 
the  other  so  as  to  coincide  in  every 
part. 


Ex.  338.  If  one  circle  of  a  sphere  passes  through  the 
poles  of  another  circle  of  a  sphere  the  planes  of  the  two 
circles  are  perpendicular  to  each  other. 

Ex.  339.  All  lines  tangent  to  a  sphere  from  the  same 
point  are  equal  and  touch  the  sphere  in  the  circumfer- 
ence of  a  circle  of  the  sphere. 

SuG.  Connect  the  center  of  the  sphere  with  the  given 
point  and  with  two  or  more  points  of  tangency. 


THE    SPHERE.  361 


PROPOSITION  XVIII.     THEOREM. 

542.  Two  symmetrical  isosceles  spherical  triangles 
can  be  made  to  coincide,  and  are  equal. 


Let  ABC  and  D  E  F  be  two  symmetrical  spherical 
truuiffles,  i.  e.,  let  A  B  =  J>  F,  A  C  =  D  E9  B  C  =  F  E, 
Z.  <  =  Z.  A  Z  B  =  ^_  F  and  ^  C  =  ^  E.  Also,  let 
ABC  and  DEF  be  isosceles  triangles,  i.  e.,  let  A  B  =  A  C 
and  DE  =  1) F. 

To  prove  that  the  triangles  ABC  and  DEF  are  equal. 

SUG.  1.  Compare  D  E  with  A  B.  Give  the  reason  for 
your  statement. 

SUG.  2.  Compare  DF  with  A  C. 

SUG.  3.   Compare  Z  D  with  Z.  A. 

SUG.  4.  Place  DEF  upon  A  B  C  so  that  D  E  shall 
fall  upon  AB  and  D  F  upon  A  C.  Why  can  this  be 
done? 

SUG.  5.  Where  will  E  fall  ?     Where  will  F  fall  ? 

SUG.  6.  Compare  each  part  of  DEF  with  a  corre- 
sponding part  of  A  B  C. 

Therefore 


362 


SOLID    GEOMETRY. 


543.  If  from  the  vertices  of  a  spherical  triangle  as 
poles,  arcs  of  great  circles  be  drawn, 

a  spherical  triangle  is  formed  which 
is  called  the  polar  of  the  first.  If 
from  A,  B  and  C  as  poles,  arcs  of 
great  circles  be  drawn,  a  triangle 
D E F\&  formed  which  is  the  polar 
of  the  triangle  ABC. 

If  entire  circles  be  drawn  they  will  intersect  to  form 
eight  spherical  triangles,  but  the  polar  of  the  given  tri- 
angle A  B  C  is  that  one  of  the  eight  triangles  whose 
vertices  lie  on  the  same  sides  of  the  arcs  of  the  given  tri- 
angle as  the  respectively  homologous  vertices  of  the 
given  triangle,  and  no  side  of  which  is  greater  than  180 
degrees. 

PROPOSITION  XIX.     THEOREM. 

544.  If  a  spherical  triangle  D  E  F  is  the  polar  of 
another  spherical  triangle  ABC,  then  the  triangle 
A  B  C  is  the  polar  of  D  E  F. 


Liet  the  spherical  triangle  D  E  F  be  the  polar  ofABC. 

To  prove  that  A  B  C  is  the  polar  of  D  E  F. 


THE   SPHERE.  363 


SUG.  1 .  A  is  the  pole  of  what  arc  ? 
SUG.  2.   How  many  degrees  from  A  to 
SUG.  3.   C  is  the  pole  of  what  arc  ? 
SUG.  4.   How  many  degrees  from  C  to  El 
SUG.  5.  E  is  the  pole  of  the  arc  A  C.     Why  ? 
SUG.  6.  F  is  the  pole  of  the  arc  A  B.     Why  ? 
SUG.  7.  D  is  the  pole  of  the  arc  B  C.     Why  ? 
Therefore 


Ex.  340.  A  sphere  may  be  inscribed  in  any  tetrahe- 
dron. 

Ex.  341.  One,  and  only  one,  surface  of  a  sphere  may 
be  described  through  any  four 
points  not  in  .the  same  plane. 

SUG.  1.  Let  A,  B,  C  and  D  be 
the  four  given  points. 

SUG.  2.  What  is  the  locus  of 
points  equally  distant  from  A 
and  #? 

SUG.  3.  What  is  the  locus  of  points  equally  distant 
from  B  and  C? 

SUG.  4.  What  is  the  intersection  of  these  two  loci  ? 

SUG.  5.  What  is  the  locus  of  points  equally  distant 
from  CandZ>? 

SUG.  6.  Does  this  last  locus  intersect  the  one  referred 
to  in  Sug.  4  ? 

Ex.  342.  A  straight  line  tangent  to  a  circle  of  a  sphere 
lies  in  a  plane  which  is  tangent  to  the  sphere  at  the 
point  of  contact. 

Ex.  343.  A  plane  tangent  to  a  sphere  is  perpendicular 
to  the  radius  through  the  point  of  contact. 


SOLID    GEOMETRY. 


PROPOSITION  XX.     THEOREM. 

545.  In  two  polar  triangles  each  angle  of  one  is 
measured  by  the  supplement  of  the  side  opposite  it  in 
tine  other. 


Let  ABC  and  D  E  F  be  two  polar  triangles. 

To  prove  that  the  angle  D  is  measured  by  the  supple- 
ment of  B  C  ? 

SUG.  1.  Extend  D  E  and  D  F  until  they  intersect  B  C 
in  M  and  N  respectively. 

SUG.  2.  B  is  the  pole  of  what  arc  ? 

SUG.  3.   How  many  degrees  in  B  Nt 

SUG.  4.   C  is  the  pole  of  what  arc? 

SUG.  5.   How  many  degrees  in  C  Ml 

SUG.  6.  How  many  degrees  in  B  N  +  M  C?  How 
many  in  BC+  M  Nt 

SUG.  7.  What  is  the  supplement  of  B  C? 

SUG.  8.  What  is  the  measure  of  Z.  D  ?     Why  ? 

Therefore 


THE    SPHERE. 


365 


PROPOSITION  XXI.     THEOREM. 

546.   Two  triangles  on  the  same  sphere,  or  equal 
spheres,  having  two  sides  and  the  included,  angle  of 
one  equal  respectively  to  two  sides  and  the  included 
angle  of  the  other,  are  either  equal  or  symmetrical* 
AD  A 


C  F  c 

Let  ABC  and  D  E  F  be  two  spherical  triangles  in 

which  the  angle  D  equals  tlie  angle  A,  and  the  sides  A  B 

and  A  C  are  equal  respectively  to  the  sides  D  E  and  D  F. 

To  prove  that  the  triangles  ABC  and  D  E  F  are  either 
equal  or  symmetrical. 

CASE  I.    When  the  parts  are  arranged  in  the  same  order. 

SUG.  1.  Place  the  A  D  E  F  upon  the  A  A  B  C,  so 
that  D  shall  fall  upon  A,  D  F  upon  A  C,  and  D  E  upon 
A  B.  Why  can  this  be  done  ? 

SUG.  2.   Where  will  F  fall  ?     Where  will  E  fall  ? 

SuG.  3.  Then  how  are  the  As  A  B  C  and  DEF  related? 

SUG.  4.  What  is  the  conclusion  in  Case  I  ? 

CASE  II.    When  the  parts  are  arranged  in  reverse  order. 

SuG.  1.  Construct  A  A  £  Cthe  symmetrical  of  A  A  B  C 
(Fig.  at  right).  Then,  since  two  sides  and  the  included 
Z.  of  DEF  are  =  to  two  sides  and  the  included  Z!  of 
ABC,  each  to  each,  but  arranged  in  the  reverse  order, 
how  do  the  parts  of  D  E  F  compare  with  those  of  A  E  C, 
each  to  each  ? 

SUG.  2.  Compare  A  ^^Cwith  A  DEF.     (Case  I.) 

SuG.  3.  Since  A  A  E  C  is  the  symmetrical  of  A  B  Ct 
how  do  As  ABC  and  DEF  compare  ? 

SUG.  4.  What  is  the  conclusion  in  Case  II  ? 

Therefore 


366 


SOLID    GEOMETRY. 


PROPOSITION  XXII.    THEOREM. 

547.  Two  triangles  on  the  same  sphere,  or  equal 
spheres,  having  two  angles  and  the  included  side  of 
one  equal  respectively  to  two  angles  and  the  included 
side  of  the  other,  are  equal  or  symmetrical. 

SUG.  See  figure  and  suggestions  of  Prop.  XXI. 
Therefore 

PROPOSITION  XXIII.    THEOREM. 

548.  Two  triangles  on  the  same  sphere,  or  equal 
spheres,  having  the  three  sides  of  one  respectively 
equal  to  the  three  sides  of  the  other,  are  either  equal 
or  symmetrical. 

A  D 


Let  ABC  and  D  E  F  be  two  spherical  triangles  on  the 
same  spJiere,  or  equal  spJieres,  having  the  three  sides  of  one 
equal  respectively  to  the  three  sides  of  the  other. 

To  prove  that  ABC  and  DBF  are  either  equal  or  sym- 
metrical. 

SUG.  1.  Connect  the  vertices  of  each  A  with  the  cen- 
ter of  the  sphere  on  which  the  A  is  situated. 

SUG.  2.  Compare  the  face  Z.§  of  the  trihedral  ^.s  O 
and  5.  (Art.  144.) 


THE    SPHERE.  367 


SUG.  3.  Compare  the  dihedral  Zs  of  the  two  trihedral 
Zs.  (Art.  390.) 

SUG.  4.  Compare  the  Zs  of  one  A  with  the  corre- 
sponding Zs  of  the  other  A.  (Art.  534.) 

SUG.  5.  Compare  the  A  A  B  C  with  the  A  DBF, 
first,  when  the  parts  are  arranged  in  the  same  order,  and 
second,  when  they  are  arranged  in  reverse  order. 

Therefore 

PROPOSITION  XXIV.    THEOREM. 

549,  Two  triangles  on  the  same  sphere,  or  equal 
spheres,  having  the  three  angles  of  one  respectively 
equal  to  the  three  angles  of  the  other,  are  equal  or  sym- 
metrical. 


Let  A  and  B  represent  two  spherical  triangles  on  the 
same  sphere,  or  equal  splieres,  having  the  three  angles  of 
one  equal  respectively  to  tJie  three  angles  of  the  other. 

To  prove  that  A  and  B  are  either  equal  or  symmetrical. 

SUG.  1.  lyet  C  and  D  be  the  polar  As  of  A  and  B  re- 
spectively. 

SUG.  2.  How  do  the  sides  of  C  and  D  compare  ?  (Art. 
545.) 

SUG  3.  How  do  the  Zs  of  C  and  D  compare  ?  (Art. 
548.) 

SUG.  4.  Then  how  do  the  sides  of  A  and  B  compare  ? 

SUG.  5.  How  are  the  As  A  and  B  related  ? 

Therefore 


308  SOLID    GEOMETRY. 

550.  A  convex   spherical   polygon   is   a   spherical 
polygon,  no    side   of  which,  if   extended,  would   pass 
within  the  polygon. 

PROPOSITION  XXV.     THEOREM. 

551.  The  sum  of  the  sides  of  a  convex  spherical 
polygon  is  less  than  the  circumference  of  a  great  circle. 

SUG.  1.  Connect  each  vertex  of  the  polygon  with  the 
center  of  the  sphere,  thus  forming  a  polyhedral  Z.  . 

SUG.  2.  Compare  the  sum  of  the  face  <</s  of  the  poly- 
hedral /_  with  four  rt.  ^/s. 

SUG.  3.  Compare  the  sum  of  the  sides  of  the  spherical 
polygon  with  the  circumference  of  a  great  O. 

Therefore 

552.  SCHOLIUM.     The  sides  of  a  spherical  polygon 
are  usually  expressed  in  degrees.     Hence  the  last  the- 
orem may  be  stated  thus:    The  sum  of  the  sides  of  a 
convex  spherical  polygon  is  less  than  four  right  angles. 


Ex.  344.  Any  side  of  a  spherical  triangle  is  less  than 
the  sum  of  the  other  two. 

SUG.  See  Art.  388. 

Ex.  345.  The  arc  of  a  great  circle  drawn  from  the  vertex 
of  an  isosceles  spherical  triangle  to  the  middle  point  of 
the  base  is  perpendicular  to  the  base  and  bisects  the  ver- 
tical angle. 

Ex.  346.  The  angles  opposite  the  equal  sides  of  an 
isosceles  spherical  triangle  are  equal. 

Ex.  347.  If  two  angles  of  a  spherical  triangle  are 
equal  the  triangle  is  isosceles. 

SUG.  Take  the  polar  of  the  given  A. 


THE    SPHERE.  369 


PROPOSITION  XXVI.    THEOREM. 

553.  The  sum  of  the  angles  of  a  spherical  triangle 
is  greater  than  two  and  less  than  six  right  angles. 


~Let  ABC  represent  a  spherical  triangle. 

To  prove  that  the  sum  of  the  angles  A ,  B  and  C  is  greater 
than  two  and  less  than  six  right  angles. 

SUG.  1.  Let  DBF  represent  the  polar  A  of  A  B  C. 
Then  Z  A  =  180°  -  E  F.  Why  ? 

SUG.  2.  Find  Zs  B  and  Cin  terms  of  D  F  and  D  E. 

SUG.  3.  Express  the  sum  of  the  Zs  A,  B  and  C  in  the 
simplest  form. 

SUG.  4.  Since  EF  +  D  F+  D  E  is  less  than  360°, 
what  may  be  said  of  the  sum  of  the  Zs  A,  B  and  C? 

Therefore 

554.  ScHOiviUM.  A  spherical  triangle,  unlike  a  plane 
triangle,  may  have  two  or  three  right  angles,  or  two  or 
three  obtuse  angles.  The  sum  of  the  angles  of  a  spheri- 
cal triangle  may  approach  six  right  angles  as  a  limit, 
and  the  sum  of  the  sides  of  a  spherical  triangle  may  ap- 
proach the  circumference  of  a  great  circle  as  a  limit. 
24— Geo. 


370 


SOLID    GEOMETRY. 


PROPOSITION  XXVII.     THEOREM. 

555.  Two    symmetrical    spherical    triangles    are 
equal  in  area. 


Let  ABC  and  D  EF  represent  two  symmetrical  spher- 
ical triangles,  i.  e.,  let  A  B  =  D  F,  A  C  =  D  E,  B  C  =  FE, 
s/_A  =  Z.I>>Z.B=Z.F  and  /_  C  =  Z  E. 

To  prove  that  the  triangles  ABC  and  DBF  are  equal 
in  area. 

SUG.  1.  Let  O  be  the  pole  of  a  small  O  through  the 
points  A,  B,  and  C.  Draw  the  arcs  of  great  Os  O  A, 
OB,  and  OC. 

SUG.  2.  Compare  the  arcs  O  A,  OB,  and  O  C . 

SUG.  3.  At  D,  draw  D  S,  making  Z.  FD  S  =  Z.  B  A  O, 
and  take  DS  =  A  O.  Draw  the  arcs  of  .great  Os  SF 
and  SE. 

SUG.  4.  In  the  spherical  As  F  D  S  and  B  A  O,  corn- 
pare  D  F  with  A  B,  DS  with  A  O,  and  Z  FD  S  with 
Z  B  A  O.  Compare  the  As  FD  S  and  B  A  O.  (Art. 
546.)  Compare  SF  with  O  B. 

SUG.  5.  DSF  and  A  OB  are  symmetrical  isosceles 
spherical  As.  Why  ?  Compare  these  As  in  respect  to 
area.  (Art.  542.) 

SUG.  6.  In  a  similar  manner,  compare  the  spherical 
As  D  S  E  and  A  O  C  in  respect  to  area. 


THE   SPHERE.  371 


SUG.  7.  Compare  the  spherical  As  ESF  and  COB 
in  respect  to  area. 

SuG.  8.  Compare  the  spherical  As  DEF  and  ABC 
in  respect  to  area. 

Therefore 

556.  A  lune  is  a  portion  of  the  surface  of  sphere  a  in- 
cluded between  two  semi-circumferences  ot         A 
great  circles,  as  A  B  D  C . 


557.  The  angle  of  a  lune  is  the  angle 
between  the  two  semi-circumferences  which 
bound  the  lune. 


Ex.  348.  Prove  Prop.  XXVII  when  the  pole  of  the 
small  circle  through  A,  B,  and  Cis  without  the  triangle. 

Ex.  349.  If  the  area  of  the  convex  surface  of  a  right 
circular  cone  is  twice  the  area  of  its  base,  prove  that  the 
slant  height  of  the  cone  is  equal  to  the  diameter  of  its 
base. 

Ex.  350.  The  radius  of  the  base  of  a  right  circular 
cone  is  5  in.,  and  the  number  of  square  inches  in  the 
area  of  the  convex  surface  of  the  cone  is  equal  to  the 
number  of  cubic  inches  in  the  volume  of  the  cone.  Find 
the  altitude  and  the  slant  height  of  the  cone. 

Ex.  351.  Which  generates  the  greater  volume,  the 
revolution  of  a  rectangle  about  its  shorter  or  its  longer 
side? 

Ex.  352.  What  is  the  locus  of  a  point  in  space  the 
sum  of  the  squares  of  whose  distances  from  two  fixed 
points  is  equal  to  the  square  of  the  distance  between  the 
two  points. 


372  SOLID    GEOMETRY. 

PROPOSITION  XXVIII.     THEOREM. 

558.  If  two  arcs  of  great  circles  intersect  on  the  sur- 
face of  a  hemisphere,  the  sum  of  the  areas  of  the  oppo- 
site spherical  triangles  thus  formed  is  equal  to  the 
area  of  a  lune  whose  angle  is  the  angle  of  the  intersect- 
ing arcs. 

A  ^— •— ^ 

B 


Let  the  arcs  of  great  circles  ACE  and  BCD  in- 
tersect at  the  point  C  on  the  hemisphere  ABED,  thus 
forming  the  opposite  spherical  triangles  ACB  and  D  CE. 

To  prove  that  the  sum  of  the  areas  of  the  triangles  ACB 
and  D  C  E  is  equal  to  the  area  of  a  lune  whose  angle  is  equal 
to  the  angle  ACB. 

SUG.  1.  Extend  the  arcs  ACE  and  B  CD  to  complete 
the  great  Os  A  CEF  and  B  CDF. 

SUG.  2.   CDFE  is  a  lune  whose  A  \$DCE.     Why  ? 

SUG.  3.  Compare  the  arcs  C  B  and  D  F\  also  the  arcs 
A  C  and  EF\  also  the  arcs  A  B  and  DE. 

SuG.  4.  Compare  the  As  A  CB  and  EFD  in  respect 
to  area.  (Arts.  548  and  555.) 

SUG.  5.  Compare  the  sum  of  the  areas  of  the  As  ABC 
and  CD  E  with  the  area  of  the  lune  CDFE. 

Therefore 


THE    SPHERE.  373 


EXERCISES. 

353.  What  is  the  locus  of  a  point  in  space  which  is  at 
a  given  distance  from  a  fixed  straight  line  ? 

354.  To  cut  a  cylinder  of  revolution  by  a  plane  paral- 
lel to  an  element  in  such  a  manner  that  the  section  shall 
be  a  rectangle   equal  in  all    respects   to  "the    rectangle 
which  generates  the  cylinder. 

355.  A  right  circular  cylinder  whose  altitude  is  5  ft. 
and  the  radius  of  whose  base  is  3  ft.  is  cut  by  a  plane 
parallel  to  the  base  at  such  a  distance  from  the  base  that 
the  area  of  the  section  is  a  mean  proportional  between  the 
areas  of  the  convex  surfaces  of  the  two  parts  into  which 
the  cylinder  is  divided.     Determine  the  lengths  of  the 
segments  of  the  altitude. 

356.  Prove  that  the  total  surface  of  a  right  circular 
cylinder  is  equal  to  the  circumference  of  its  base  multi- 
plied by  the  sum  of  its  altitude  and  the  radius  of  its 
base. 

357.  If  four  similar  cylinders  of  revolution  have  their 
altitudes  proportional  to  the  numbers  3,  4,  5  and  6,  prove 
that  the  volume  of  the  largest  cylinder  is  equal  to  the 
sum  of  the  volumes  of  the  other  three. 

358.  If  two  regular  tetrahedrons  of  different  volumes 
have  the  edges  of  one  parallel  respectively  to' those  of  the 
other,  prove  that  the  lines  joining  corresponding  vertices 
meet  in  the  same  point. 

359.  The  dimensions  of  a  rectangular  parallelepiped 
are  3,  4  and  12  ft.     Find  the  circumference  of  a  great 
circle  of  the  circumscribing  sphere. 

360.  What  is  the  volume  of  a  piece  of  timber  15  ft. 
long,  whose  bases  are  squares,  each  side  of  one  base  be- 
ing 14  in.,  and  each  side  of  the>  other  base  12  in  ? 


374 


SOLID   GEOMETRY. 


SPHERICAL,  AREAS. 

559.  A  spherical  degree,  or  a  degree  of  spherical 
surface,  is  one  three  hundred  and  six- 
tieth of  the  surface  of  a  hemisphere. 

560.  If  three  great  circles  are  per- 
pendicular to  one  another,  eight  spher- 
ical triangles  are  formed,  each  having 
three  right  angles.     Bach  of  these  tri- 
angles is  called  a  tri-rectangular  triangle. 

A  tri-rectangular  triangle  has  ninety  spherical  degrees. 

PROPOSITION  XXIX.     THEOREM. 

561.  The  area  of  a  lune  is  to  the  area  of  the  surface 
of  a  sphere  as  the  angle  of  the  lune  is  to  four  right 
angles. 

A 


Let  A  J5  D  C  represent  a  lune  tvhose  angle  is  B  A  C. 

To  prove  that  the  area  of  A  B  D  C  is  to  the  area  of  the 
surface  of  the  sphere  as  the  angle  BAG  is  to  four  right 
angles. 

Let  B  C  be  an  arc  of  a  great  O  whose  poles  are  A  and 
D,  and  let  O  be  the  center  of  the  sphere. 


THE   SPHERE.  375 


CASK  I.  When  Z  B  O  C  and  four  rt.  Z.s  are  commen- 
surable, 

SUG.  1.  Draw  lines  dividing  the  Z.  B  O  C  and  4  rt. 
Z.s  into  equal  parts,  each  of  which  is  a  common  unit  of 
measure  of  Z.  B  O  C  and  4  rt.  Z^s.  Let  this  unit  be 
contained  m  times  in  Z.  B  O  C,  and  n  times  in  4  rt.  Z!s. 

SUG.  2.  What  is  the  ratio  -Z  B4  °  *T  equal  to? 

4.  rt.  Z_s 

SUG.  3.  Pass  a  plane  through  the  lines  of  division  of  the 
Z.s  and  the  edge  of  the  lune  A  D.  These  planes  divide 
the  surface  of  the  sphere  into  small  lunes. 

SUG.  4.  Compare  these  small  lunes  in  respect  to  area. 

SUG.  5.  How  does  the  number  of  lunes  compare  with 
the  number  of  angles  ? 

surface  of  lune 

SUG.  6.   What  is  the  ratio  -  —  equal  to  ? 

surface  of  sphere 

SUG.  7.  Compare  answers  to  Sugs.  2  and  6.  What  is 
the  conclusion  when  Z.  B  O  C  and  4  rt.  ^s  are  commen- 
surable ? 

CASK  II.  When  Z.  BOC  and  four  rt.  Z.s  are  incommen- 
surable. 

SUG.  Employ  the  method  of  limits  similar  to  Arts. 
171,  254,  424,  etc. 

Therefore 

562.  COROLLARY.  A  lune  contains  twice  as  many 
spherical  degrees  as  its  angle  contains  angular  degrees. 

Let  5  represent  the  number  of  spherical  degrees  in  the 
surface  of  the  lune,  and  A  the  number  of  angular  degrees 
in  the  Z.  of  the  lune.  Then,  as  there  are  720  spherical 
degrees  in  the  surface  of  a  sphere,  it  is  evident  from  the 
proposition  that 

5  A 

720  =  :    360  * 

Hence  S=<2A. 


376  SOLID  GEOMETRY. 

563.  The  spherical  excess  of  a  spherical  triangle  is 
the  excess  of  the  sum  of  its  angles  over  two  right  angles. 

PROPOSITION  XXX.     THEOREM. 

564.  The  number  of  spherical  degrees  in  a  spheri- 
cal triangle  is  equal  to  the  n.uiriber  of  angular  degrees 
in  its  spherical  excess. 


Let  A  It  C  be  a  spherical  triangle. 

To  prove  that  the  number  of  spherical  degrees  in  ABC 
is  equal  to  the  mimber  of  angular  degrees  in  ^_  A  -f  ^_  B 
+  Z.C-  180°. 

SUG.  1.  Let  one  side  of  the  A,  as  B  C,  be  extended 
to  form  a  complete  great  0  B  C  D  E.  Let  B  A  and 
C  A  be  extended  to  meet  the  great  0  B  C  D  E  in 
D  and  E  respectively. 

SUG.  2.  BCD,BADtm&BED  are  semi-  circum- 
ferences of  great  Os.  Why  ? 

SUG.  3.  &ABC+&A  ED  =  lune  whose  Z  is  A 
Why? 

SUG.  4.  &A  B  C+  &A  CD=  lune  whose  Z  is  B 
Why? 

SUG.  5.  &ABC+&ABE  =  lune  whose  Z  is  C 
Why? 


THE    SPHERE.  377 


SUG.  6.  How  many  spherical  degrees  in  a  lune  whose 
^  is  A  ?  in  a  lune  whose  ^  is  B  ?  in  a  lune  whose  /_ 
is  C?  Give  auth. 

SuG.  7.   How  many  spherical  degrees  in  a  hemisphere  ? 

SUG.  8.  If  the  number  of  spherical  degrees  in  A  A  B  C 
be  represented  by  m,  then,  from  the  equations  in  Sugs. 
3,  4  and  5,  2  m  +  360  equals  what  ? 

SUG.  8.  What  does  w  equal  ? 

Therefore 

565.  SCHOLIUM.  To  say  that  a  spherical  triangle 
contains  a  certain  number,  say  n,  of  the  spherical  de- 
grees, is  simply  to  say  that  the  surface  of  the  triangle  is 

equal  to  -~  of  the  surface  of  a  hemisphere,  or  to 


of  the  surface  of  a  sphere.  The  whole  surface  of  a  sphere, 
expressed  in  square  measure,  will  be  found  in  proposition 
XXXIII,  and  then  the  area  of  the  surface  of  a  spherical 
triangle  can  be  expressed  in  square  measure,  by  means 
of  the  theorem  just  given,  when  the  angles  of  the  tri- 
angle and  the  radius  of  the  sphere  are  known. 


Ex.  361.  How  many  spherical  degrees  are  there  in  a 
spherical  triangle  whose  angles  are  200°,  140°  and  100°  ? 

Ex.  362.  What  part  of  the  surface  of  a  sphere  is  a 
spherical  triangle  whose  angles  are  120°,  140°,  160°  ? 

Ex.  363.  What  part  of  the  surface  of  a  sphere  is  a 
spherical  triangle  each  angle  of  which  is  90°  ?  How 
many  spherical  degrees  in  the  same  triangle  ? 

Ex.  364.  Find  the  angles  of  an  equiangular  spherical 
triangle  whose  surface  is  y1^  of  the  surface  of  a  sphere. 


378  SOLID    GEOMETRY. 

566.  A  zone  is  that  portion  of  the  surface  of  a  sphere 
which  is  included  between  two  parallel  planes. 

The  circumferences  of  the  sections  of  the  sphere  made 
by  the  planes  are  called  the  bases  of  the  zone,  and  the 
perpendicular  distance  between  the  parallel  planes  is 
called  the  altitude  of  the  zone. 

567.  A  spherical  segment  is  a  portion  of  a  sphere 
included  between  two  parallel  planes. 

The  sections  of  the  sphere  by  the  planes  are  called  the 
bases  of  the  segment,  and  the  perpendicular  distance  be- 
tween the  planes  is  called  the  altitude  of  the  segment. 

A  zone  is  the  spherical  surface  of  a  segment. 

The  portion  of  a  sphere  cut  off  by  any  plane  is  considered  as  a 
spherical  segment  for  it  is  included  between  the  cutting  plane  and  a 
plane  tangent  to  the  sphere  parallel  to  the  cutting  plane.  For  the 
same  reason,  the  portion  of  the  surface  of  a  sphere  cut  off  by  any 
plane  is  considered  as  a  zone.  In  this  case  the  segment  and  zone. are 
are  said  to  be  of  one  base. 

568.  If  a  semi-circle  be  revolved  about 
its  diameter  as  an  axis,  a  sphere  will  be 
generated;   any  arc  of  the  semi-circum- 
ference will  generate  a  zone. 


569.  A  spherical  sector  is  the  por- 
tion of  a  sphere  generated  by  the  revolu-        B 
tion  of  a  circular  sector  about  a  diameter. 

The  pupil  should  form  a  mental  picture  of  the  different  varieties 
of  spherical  sectors  and  describe  them.  For  example,  if  the  semi- 
circle A  D  B  be  revolved  about  the  diameter,  the  circular  sector,  AOC, 
will  generate  a  spherical  sector  whose  surface  is  a  zone  of  one  base 
generated  by  the  arc  A  C,  and  a  convex  conical  surface  generated  by 
the  radius  O  C.  The  circular  sector  COB  generates  a  spherical 
sector  whose  surface  is  a  zone  of  one  base  generated  by  the  arc  C  D  B 
and  a  concave  conical  surface  generated  by  the  radius  O  C. 


THE   SPHERE.  37$ 


EXERCISES. 

365.  Construct  a  semi-circle,  and  in  it  a  circular  sector 
which,  if  revolved,  will  generate  a  spherical  sector  hav- 
ing two  concave  conical  surfaces.     Describe  the  zone. 

366.  Construct  a  circular  sector  which  will  generate  a 
spherical  sector  whose  surface  will  be  a  concave  conical 
surface,  a  plane  surface,  and  a  zone. 

367.  Is  a  hemisphere  a  spherical  sector? 

368.  What  is  the  locus  of  a  point  in  space  which  is  at 
a  given  distance,  a,  from  a  given  plane,  and  at  a  given 
distance  greater  than  a  from  a  given  point  in  the  given 
plane  ? 

Ex.  369.  Compute  the  lateral  area,  the  total  area,  and 
the  volume  of  the  frustum  of  a  cone  of  revolution,  given 
the  altitude  of  the  cone,  20  ft.;  the  diameter  of  the  lower 
base,  16  ft.;  the  diameter  of  the  upper  base,  12  ft. 

370.  A  cylinder  of  revolution,  and  the  frustum  of  a 
cone  of  revolution,  have  the  same  lower  base  and  the 
same  altitude;  what  must  be  the  ratio  of  the  radii  of  the 
two  bases  of  the  frustum  in  order  that  the  volume  of  the 
frustum  may  be  one  half  the  volume  of  the  cylinder  ? 

371.  If  a  rectangle  revolves  about  one  of  its  sides  the 
volume  generated  is  288  ft  cubic  ft.,  but  if  it  revolves 
about  the  adjacent  side  the  volume  generated  is  384  n 
cubit  ft;  find  the  diagonal  of  the  rectangle. 

372.  A  regular  quadrangular  pyramid  has  each  basal 
edge  equal  to  12  ft.,  and  each  lateral  edge  equal  to  10  ft. ; 
find  the  altitude  of  the  pyramid. 

,  373.  Find  the  radius  of  the  base  of  a  cylinder  of  revo- 
lution whose  altitude  is  1  yd.,  and  whose  volume  is  48  n 
cubic  ft. 


•380 


SOLID    GEOMETRY. 


PROPOSITION  XXXI.     THEOREM. 

570.  The  area  of  the  surf  ace  generated  by  the  revo- 
lution of  a  straight  line  about  an  axis  in  its  plane, 
but  not  crossing  it,  is  equal  to  the  projection  of  the  line 
upon  the  ajcis  multiplied  by  the  circumference  of  a 
circle  whose  radius  is  a  perpendicular  erected  at  the 
middle  point  of  the  line  and  limited  by  the  axis. 


Let  H  S  be  the  axis  of  revolution,  A  B  the  line  which 
revolves  to  generate  the  surface,  CD  the  projection  of  AB 
upon  K  S9  M  the  middle  point  of  A  B,  and  M  O  the  per- 
pendicular to  the  line  A  B,  at  M. 

To  prove  that  the  surface  generated  by  A  B  is  equal  to 
C  D  multiplied  by  the  circumference  of  a  circle  whose  radius 
is  M  O. 

CASE  I.      When  A  B  does  not  meet  the  axis. 

SUG.  1.  A  B  generates  the  convex  surface  of  the  frus- 
tum of  a  cone  of  revolution.  What  is  the  area  of  this 
surface  equal  to?  (Art.  504.) 

SUG.  2.  Draw  A  C,  MN,  and  B  D  _L  to  R  5,  and  A  P 
II  to  R  S. 


THE    SPHERE.  381 


SUG.  3.  Prove  that  MN  equals  one  half  otAC  +  BD, 
and  hence  the  circumference  of  a  0  generated  by  M 
equals  one  half  the  sum  of  the  circumferences  of  the  up- 
per and  lower  bases  of  the  frustum. 

SUG.  4.  The  As  MNO  and  A  PB  are  similar.   Why  ? 

Compare  the  ratio  -TT    with  -TFTT  and  hence  with  -rrrf- 


SUG.  5.  I^et  a  be  the  measure  of  A  B,  c  the  measure  of 
C  D,  r  the  measure  of  M  O,  and  n  the  measure  of  M  N. 

SUG.  6.  Compare  the  ratios  —  and  — 

r  n 

SUG.  7.  Compare  a  n  and  c  r. 

SUG.  8.  The  measure  of  the  surface  generated  by  A  B 
equals  2  n  a  n.  Why  ?  It  also  equals  2  n  cr.  Why  ? 

SUG.  9.  What  is  the  circumference  of  a  O  whose  ra- 
dius is  r  equal  to  ?  Hence,  what  is  the  area  of  the  sur- 
face generated  by  A  B  ? 

CASE  II.      When  one  point,  as  A,  is  on  the  axis. 

SUG.  This  case  is  obtained  from  Case  I,  by  making 
A  C  equal  to  nothing.  Examine  the  reasoning  in  Case 
I,  and  see  that  it  holds  when  A  C  =  0. 

CASK  III.      When  A  B  is  parallel  to  the  axis. 

SUG.  This  case  is  obtained  from  Case  I,  by  making 
A  C  =  B  D.  Examine  the  reasoning  in  Case  I  and  see 
that  it  holds  when  A  C  =  B  D. 

Therefore  - 

QUERY.  What  is  the  surface  generated  by  A  B  called 
when  A  is  on  the  axis  ?  What  when  A  B  is  parallel  to 
the  axis  ? 


382 


SOLID    GEOMETRY. 


PROPOSITION  XXXII.     THEOREM. 

571.  The  area  of  a  zone  equals  the  product  of  its 
altitude  by  the  circumference  of  a  great  circle. 


luet  XCT  represent  a  semi-circle,  and  A  D  an  arc 
which,  revolved  around  X  Y,  generates  a  zone. 

To  prove  that  the  area  of  the  zo?ie  is  equal  to  its  altitude 
multiplied  by  the  circumference  of  a  great  circle. 

SUG.  1.  Divide  the  arc  A  D  into  any  number  of  equal 
parts,  as  A  B>  B  C,  etc.,  and  draw  the  chords  of  these 
arcs. 

SuG.  2.  How  do  these  chords  compare  in  length  ? 

SUG.  3.  At  the  middle  points  of  these  chords  erect  J_s 
terminating  in  the  axis  X  Y. 

SUG.  4.  Where  will  these  J_s  intersect  ?     Why  ? 

SUG.  5.   How  do  these  J_s  compare  in  length  ?   Why  ? 

SUG.  6.  What  is  the  area  generated  by  the  chord  A  B  ? 
by  the  chord  B  C?  etc. 

SUG.  7.  Express,  in  its  simplest  form,  the  area  gener- 
ated by  all  these  chords  taken  together. 

SUG.  8.  Bisect  the  arcs,  draw  chords  to  the  points 
of  bisection,  and  proceed  as  before. 


THE    SPHERE.  383 


SUG.  8.  Let  the  number  of  equal  parts  into  which  the 
given  arc  is  divided  be  continually  increased,  then  the 
surface  generated  by  the  chords  of  these  arcs  taken  to- 
gether is  a  variable.  But  during  its  variation  it  is  al- 
ways equal  to  what  ? 

SUG.  9.  What  two  variables  are  here  always  equal  ? 

SUG.  10.  What  are  the  respective  limits  of  these  va- 
riables ? 

SUG.  11.  What  is  the  area  of  the  zone? 

Therefore  - 

572.  COROLLARY.  The  above  theorem  may  be  ex- 
pressed by  the  formula 


in  which  5  represents  the  area  of  the  surface  of  the  zone, 
H  its  altitude,  and  R  the  radius  of  the  sphere  on  which 
the  zone  lies. 


Ex.  374.  Prove  that  the  sum  of  the  angles  of  a  convex 
spherical  polygon  of  n  sides  is  greater  than  2  n  —  4  and 
less  than  6  n  —  12  right  angles. 

SUG.  Divide  the  spherical  polygon  of  n  side  into  n  —  2 
spherical  triangles  by  arcs  of  great  circles  drawn  from  one 
vertex  to  each  non-adjacent  vertex. 

Ex.  375.  Prove  that  the  number  of  spherical  degrees 
in  a  convex  spherical  polygon  of  n  sides  is  equal  to  the 
number  which  expresses  the  excess  of  the  sum  of  its 
angles  over  2  n  —  4  right  angles. 

See  Sug.  in  Ex.  374. 

Ex.  376.  The  volume  of  a  rectangular  parallelepiped  is 
6,720  cubic  inches,  and  its  edges  are  in  the  ratio  of  the 
numbers  3,  5  and  7;  find  the  three  ed°;es. 


384  SOLID    GEOMETRY. 

PROPOSITION  XXXIII.     THEOREM. 

573.  The  area  of  the  surface  of  a  sphere  equals  the 
product  of  its  diameter  by  the  circumference  of  a 
great  circle. 

SUG.  Employ  the  method  of  Prop.  XXXII,  or  use 
that  proposition  as  authority,  taking  a  semi-circumfer- 
ence as  the  given  arc. 

574.  COROW,ARY  I.     The  trutn.  of  the  theorem  may 
be  expressed  by  the  formula 

S=%7tRD,  or,  5=477- .£2, 

in  which  £  represents  the  surface  of  the  sphere,  R  its 
radius,  and  D  its  diameter. 

575.  CORONARY  II.  •  The  area  of  the  surface  of  a 
sphere  equals  the  area  of  four  of  its  great  circles.     (See 
second  formula  in  Cor.  I.) 

576.  CORONARY  III.     The  area  of  the  surface  of  a 
sphere  equals  the  area  of  a  circle  whose  radius  is  equal 
to  the  diameter  of  the  sphere.     (See  second  formula  in 
Cor.  I.) 

• 

577.  CORONARY  IV.     The  areas  of  the  surfaces  of 
two  spheres  have  the  same  ratio  as  the  squares  of  their 
radii  or  the  squares  of  their  diameters. 

578.  COROU,ARY  V.     The  area  of  a  spherical  degree 
may  be  expressed  as     nor.  - .     (Art.  559.) 

1 2i\) 


Ex.  377.  Find,  in  square  feet,  the  area  of  a  spherical 
triangle  whose  angles  are  95°,  140°  and  216°,  the  radius 
of  the  sphere  being  15  in. 


THE    SPHERE. 


385- 


PROPOSITION  XXXIV.     THEOREM. 

579.  The  volume  of  a  sphere  is  equal  to  the  area  of 
its  surface  multiplied  by  one  third  of  its  radius. 


SUG.  1  .  Circumscribe  a  polyhedron  about  the  sphere. 

SUG.  2.  Join  each  vertex  of  the  polyhedron  with  the 
center  of  the  sphere,  and  pass  planes  through  these  lines 
and  the  edges  of  the  polyhedron.  Pyramids  are  thus 
formed.  Why  ? 

SUG.  3.  Compare  the  altitude  of  these  pyramids  with 
the  radius  of  the  sphere. 

SUG.  4.  What  is  the  volume  of  each  pyramid  equal  to  ? 

SUG.  5.  What  is  the  volume  of  the  polyhedron  ? 

SUG.  6.  Circumscribe  a  polyhedron  of  a  greater  num- 
ber of  sides  about  the  sphere.  What  is  its  volume  in 
terms  of  its  surface  ? 

SUG.  7.  Finish  the  demonstration. 

Therefore  - 

580.  COROLLARY  I.  The  volume  of  a  sphere  may  be 
expressed  by  the  formula 


in  which  V  represents  the  volume  and  R  the  radius  of 

the  sphere. 
25—  Geo. 


386  SOLID    GEOMETRY. 

581.  COROLLARY  II.     The  volumes   of  two   spheres 
have  the  same  ratio  as  the  cubes  of  their  radii,  or  the 
cubes  of  their  diameters. 

582.  A  spherical  pyramid  is  a  solid  bounded  by  a 
spherical  polygon,  and  the  planes 

of  the  sides  of  the  polygon,  as 
O-ABCDE. 

The  vertex  of  the  pyramid  is 
at  the  center  of  the  sphere.  The 
spherical  polygon  is  the  base  of 
the  spherical  pyramid. 


Ex.  378.  The  volume  of  a  spherical  pyramid  is  equal 
to  the  area  of  its  base  multiplied  by  one  third  of  the  ra- 
dius of  the  sphere. 

SUG.  Employ  the  method  of  Prop.  XXXIV. 

Ex.  379.  The  volume  of  a  spherical  sector  is  equal  to 
the  area  of  its  zone  multiplied  by  one  third  of  the  radius 
of  the  sphere. 

SUG.  Employ  the  method  of  Prop.  XXXIV. 

Ex.  380.  Prove  that  the  volume  of  a  sphere  is  twice 
the  volume  of  a  cone  whose  altitude  is  equal  to  the  diam- 
eter of  the  sphere,  and  the  radius  of  whose  base  is  equal 
to  the  radius  of  the  sphere. 

Ex.  381.  The  radius  of  a  sphere  is  3  in.,  and  the  area 
of  a  spherical  triangle  A  B  C  on  this  sphere  is  11.7  it. 
The  angles  A  and  B  are  72°  and  115°  respectively.  Find 
the  angle  C  of  the  spherical  triangle. 

Ex.  382.  On  the  same  sphere,  or  equal  spheres,  zones 
of  equal  altitudes  are  equal  in  area. 


THE    SPHERE. 


387 


PROPOSITION  XXXV.     PROBLEM. 
583.  To  find  the  volume  of  a  spherical  segment. 


Let  A  B  D  C  generate  a  spJierical  segment  by  revolving 
about  A  C. 

To  find  the  volume  of  the  segment. 

SUG.  1.  Join  B  and  D  to  the  center  of  the  semi- 
circle O. 

Sue.  2.  Find  the  volume  of  the  spherical  sector  gen- 
erated by  revolving  the  circular  sector  BOD.  (Ex.  379.) 

SUG.  3.  Add  the  volume  of  the  cone  generated  by  re- 
volving the  A  B  A  O. 

SUG.  4.  Subtract  the  volume  generated  by  revolving 
the  A  D  C  O. 


Ex.  383.  Find  the  volume  of  a  spherical  sector  having 
a  zone  of  10  in.  altitude  on  a  sphere  of  20  in.  radius. 

Ex.  384.  Find  the  volume  of  a  spherical  segment  of 
8  in.  altitude,  one  base  of  which  passes  through  the  cen- 
ter of  the  sphere,  the  radius  of  the  sphere  being 
20  in. 


388  SOLID    GEOMETRY. 


EXERCISES. 

385.  A  sphere  is  cut  by  parallel  planes  so  that  the 
diameter  is  divided  into  ten  equal  parts:  compare  the 
area  of  the  zones;   also  the  volumes  of  the   spherical 
sectors  whose  spherical  surfaces  are  the  respective  zones. 

386.  If  the  diameter  of  the  sphere  is  30  ft.,  find  the 
volumes  of  the   first,   fifth,   and  seventh   segments  re- 
ferred to  in  exercise  385. 

387.  Find  the  volume  and  area  of  the  surface  oi  the 
sphere  in  exercise  386. 

388.  The  sides  of  a  triangle,  on  a  sphere  whose  ra- 
dius is   10   ft.,   are,  respectively,   95°,    117°,    and   37°. 
Find  the  area  in  square  feet  of  its  polar  triangle. 

389.  Assuming   the  earth  to   be  a  sphere  whose  di- 
ameter is  7912  miles,  how  many  square  miles  upon  its 
surface. 

390.  If  the  average  specific  gravity  of  the  earth  is 
7£,  what  is  its  weight  expressed  in  tons. 

391.  Assuming    the    diameter    of   the    earth    to    be 
8,000  miles,  and  that  of  the  moon  2,000;  how  do  the 
amounts  of  light  reflected  from  them  to  a  point  in  space 
equally  distant  from  each  compare  ? 

392.  With  the  same   assumption  as  that  of  exercise 
391,  what  is  the  ratio  oi  the  volumes  of  the  earth  and 
moon? 

393.  A  triangle  on  a  12  in.  globe  has  for  its  angles 
140°,  119°,  and  196°  respectively;  compute  its  area. 

394.  The  diameter  of  a  sphere  is  equal  to  the  altitude 
of  a  cone  of  revolution  and  of  a  cylinder  of  revolution, 
and  the  radius  of  the  sphere  is  equal  to  the  radius  of  the 
cone  and  of  the  cylinder.     Prove  that  the  volumes  of  the 
cone,  sphere,  and  cylinder  are  proportional  to  the  num- 
bers 1,  2,  and  3. 


PROPOSITIONS  IN  CHAPTER  VIII. 


PROPOSITION  I. 

Every  section  of  a  cylinder  made  by  a  plane  containing  an  element 
is  a  parallelogram. 

PROPOSITION  II. 
The  bases  of  a  cylinder  are  equal.  . 

PROPOSITION  III. 

The  area  of  the  lateral  surface  of  a  cylinder  is  equal  to  the  perim- 
eter of  a  right  section  multiplied  by  an  element  of  the  surface. 

PROPOSITION  IV. 

The  volume  of  a  cylinder  equals  the  area  of  a  right  section  multi- 
plied by  a  lateral  edge. 

PROPOSITION  V. 

Every  section  of  a  cone  made  by  a  plane  passing  through  the  ver- 
tex is  a  triangle. 

PROPOSITION  VI. 

Every  section  of  a  circular  cone  made  by  a  plane  parallel  to  the 
base  is  a  circle. 

PROPOSITION  VII. 

The  area  of  the  convex  surface  of  a  cone  of  revolution  is  equal  to 
one  half  the  product  of  the  perimeter  of  its  base  by  its  slant  height. 

PROPOSITION  VIII. 

The  area  of  the  convex  surface  of  a  frustum  of  a  cone  of  revolution 
is  equal  to  one  half  the  product  of  its  slant  height  by  the  sum  of  the 
perimeters  of  its  bases. 

PROPOSITION  IX. 

The  volume  of  a  cone  equals  one  third  of  the  product  of  the  area 
of  its  base  by  its  altitude. 


390  SOLID   GEOMETRY. 

PROPOSITION  X. 
Every  section  of  a  sphere  made  by  a  plane  is  a  circle. 

PROPOSITION  XI. 
Two  intersecting  great  circles  of  a  sphere  bisect  each  other. 

PROPOSITION  XII. 

Three  points  on  the  surface  of  a  sphere  determine  a  circle  of  the 
sphere. 

PROPOSITION  XIII. 

The  shortest  distance  on  the  surface  of  a  sphere  between  any  two 
points  on  that  surface  is  the  arc,  not  greater  than  a  semi-circumfer- 
ence, of  a  great  circle  which  joins  them. 

PROPOSITION  XIV. 

All  points  in  the  circumference  of  a  circle  are  equally  distant  from 
either  of  its  poles. 

PROPOSITION  XV. 
Given  a  material  sphere,  to  find  its  radius. 

PROPOSITION  XVI. 

A  plane  perpendicular  to  a  radius  of  a  sphere  at  its  extremity,  is 
tangent  to  the  sphere. 

PROPOSITION  XVII. 

A  spherical  angle  is  equal  to  the  dihedral  angle  formed  by  the 
planes  of  the  two  arcs,  and  is  measured  by  the  arc  of  a  great  circle 
described  from  the  intersection  of  the  arcs  as  a  pole,  and  intercepted 
between  them. 

PROPOSITION  XVIII. 

Two  symmetrical  isosceles  spherical  triangles  can  be  made  to  co- 
incide, and  are  equal. 

PROPOSITION  XIX. 

If  a  spherical  triangle  D  E  F  is  the  polar  of  another  spherical  tri- 
angle ABC,  then  the  triangle  A  B  C  is  the  polar  of  D  E  2\ 


THE    SPHERE.  391 


PROPOSITION  XX. 

In  two  polar  triangles  each  angle  of  one  is  measured  by  the  sup- 
plement of  the  side  opposite  it  in  the  other. 

PROPOSITION  XXI. 

Two  triangles,  on  the  same  sphere,  or  equal  spheres,  having  two 
sides  and  the  included  angle  of  one  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other,  are  either  equal  or  symmetrical. 

PROPOSITION  XXII. 

Two  triangles,  on  the  same  sphere,  or  equal  spheres,  having  two 
angles  and  the  included  side  of  one  equal  respectively  to  two  angles 
and  the  included  side  of  the  other,  are  either  equal  or  symmetrical. 

PROPOSITION  XXIII. 

Two  triangles,  on  the  same  sphere  or  equal  spheres,  having  the 
three  sides  of  one  respectively  equal  to  the  three  sides  of  the  other, 
are  either  equal  or  symmetrical. 

PROPOSITION  XXIV. 

Two  triangles  on  the  same  sphere,  or  equal  spheres,  having  the 
three  angles  of  one  respectively  equal  to  the  three  angles  of  the  other, 
are  either  equal  or  symmetrical. 

PROPOSITION  XXV. 

The  sum  of  the  sides  of  a  convex  spherical  polygon  is  less  than  the 
circumference  of  a  great  circle. 

PROPOSITION  XXVI. 

The  sum  of  the  angles  of  a  spherical  triangle  is  greater  than  two 
and  less  than  six  right  angles. 

PROPOSITION  XXVII. 
Two  symmetrical  spherical  triangle  are  equal  in  area. 

PROPOSITION  XXVIII. 

If  two  arcs  of  great  circles  intersect  on  the  surface  of  a  hemisphere, 
the  sum  of  the  areas  of  the  opposite  spherical  triangles  thus  formed 
is  equal  to  the  area  of  a  lune  whose  angle  is  the  angle  of  the  intersect- 
ing arcs. 


392  SOLID    GEOMETRY. 


PROPOSITION  XXIX. 

The  area  of  a  lune  is  to  the  area  of  the  surface  of  a  sphere  as  the 
angle  of  the  lune  is  to  four  right  angles. 

PROPOSITION  XXX. 

The  number  of  spherical  degrees  in  a  spherical  triangle  is  equal  to 
the  number  of  angular  degrees  in  its  spherical  excess. 

PROPOSITION  XXXI. 

The  area  of  the  surface  generated  by  the  revolution  of  a  straight 
line  about  an  axis  in  its  plane,  but  not  crossing  it,  is  equal  to  the  pro- 
jection of  the  line  upon  the  axis  multiplied  by  the  circumference  of  a 
circle  whose  radius  is  a  perpendicular  erected  at  the  middle  point  of 
the  line  and  limited  by  the  axis. 

PROPOSITION  XXXII. 

The  area  of  a  zone  equals  the  product  of  its  altitude  by  the  cir- 
cumference of  a  great  circle. 

PROPOSITION  XXXIII. 

The  area  of  the  surface  of  a  sphere  equals  the  product  of  its  diam- 
eter by  the  circumference  of  a  great  circle. 

PROPOSITION  XXXIV. 

The  volume  of  a  sphere  is  equal  to  the  area  of  its  surface  multi- 
plied by  one  third  its  radius. 

PROPOSITION   XXXV. 
To  find  the  volume  of  a  spherical  segment. 


INDEX. 


Abbreviations,  0. 
Acute  triangle,   IS. 
Acute  angle,  (5. 
Adjacent  angles,  3. 
Alternate  exterior,    Jl. 
Alternate  interior,  41. 
Alternation.  144. 

Altitude,  <•>!>,  -js-.j.  rsn'.i.  :;io.  :;; 

Angles,    I.  :;i 

Angle  at  the  center,  21  1 
Antecedent,   KM. 
Apothem,  '211. 
Arc.  83 

Area,  1 7«i.  isc,,  374. 
Axiom,  s,    K>.  -j|2. 
Axis  of  a  circle,  :!  !'.». 

Basal  edges,  288,  :!<>!>. 

is,  <;•.». 
Broken  line,  3. 

Center,  S2.  '.Ml. 

Chor<i 

Circles,  S2. 

Circular  cylinder,  ::::<» 

Circumference,  v.1 

Circumscribed,  IK!,  117,333,343. 

Commensurable,   !<»;) 

Complement,  (5. 

Composition,  14."). 

Concave  polygon 

Concentric  circles,  129. 

Conclusion,  7,  30. 

Cones,  :;:;'.» 

Cones  of  revolution,  340. 
Conical  surface,  :!:!!). 
Conjugate  angles,  4. 
Consequent,   137. 
Constant.  10  I. 
Construction,  118. 
Converse  of  a  proposition,  30,  53. 
Convex  polygon,  68. 


Corollary,  8. 

Corresponding  angles,  41. 
Cube,  2!»;. 
Curved  line,  3. 
Cylinder,  329. 

Definition,  72. 
Degree,  33. 
Demonstration,  7. 
Diagonal,  07. 
Diameter,  82,  89,  347. 
Dihedral  angles,  2<i|. 
Directrix,  :!29. 
Distance,  58. 
Dividers,  118. 
Division,  100,  14(5. 
Dodecahedron,  I>'J  I. 

Element,  339. 
Enunciation,  7. 
Equiangular  polygon,  08. 
Equilateral  polygon,  (>S. 
Equilateral  triangle,   1  s. 
Exterior  angles,  41,  48.  (17. 
Extremes,  137. 

Fourth  proportional,  107. 
Frustum,  310,  340. 

Gauss,  238. 

General  enunciation,  7. 

Generatrix,  329. 

Geometrical  figure,  2. 

Geometry,  3. 

Great  circle  of  sphere,  349. 

Greatest  common  unit,  101. 

Hexagon,  08. 
Homologous,  152. 
Hypotenuse,  38. 
Hypothesis,  7. 


394: 


INDEX. 


Icosahedron,  325. 
Incommensurable   number,    103, 

104. 

Incommensurable  quantity,   102. 
Incommensurable  ratio,  103. 
Indirect  method,  54,  55. 
Inscribed,  112,  110,  117,  333,  343. 
Interior  angles,  41,  67. 
Inversion,  145. 
Isosceles  triangle,  18. 
Isosceles  trihedral,  28i 

Lateral  edges,  288,  309. 

Lateral  faces,  288. 

Leg,  09. 

Limits,  104. 

Line,  2. 

Locus  of  a  point,  27,  28,  261,  262. 

Logical  terms,  7. 

Lune,  371. 

Magnitudes,  3,  4,  33. 
Means,  137. 

Mean  and  extreme  ratio,  204. 
Mean  proportional,  140. 
Measurement,  99. 
Measure  of  a  quantity,  99. 
Models,  12,  15,  16,  20,  23,  32,  36, 
46,  109,  180. 

Notes: 

Area,  182. 

Base,  69. 

Circle,  82. 

How  to  study,  11. 

Indirect  demonstration,  54. 

Intersection  of  loci,  179. 

Limits,  104. 

Parallelograms,  70. 

Property  and  definition,   72. 

Propositions,   139,   140,   142, 
148. 

Ratio,  100,  179. 

Suggestions,  122. 
Number,  99,  100. 

Oblique  angle,  6. 
Obtuse  angle,  6. 
Obtuse  triangle,  18. 
Octahedron,  322. 


Opposite  angle,  6. 

Opposite  interior  angles,  49. 

Parallelogram,  68. 

Parallel,  40,  254. 

Parallelepiped,  297. 

Pentagon,  68. 

Pentahedron,  287. 

Perimeter,  67. 

Perpendicular,  6,  246. 

Pi  (if),  225,  2->8,  2:5:1. 

Plane  angle  of  a  dihedral,  264. 

Plane,  or  plane  surface,  3. 

Plane  figure,  .'5. 

Plane  geometry,  3. 

Plane  triangle,  18. 

Point,  2. 

Point  of  tangency,  83. 

Polar  distance,  35(5. 

Polar  triangle,  362. 

Poles  of  a  circle,  349. 

Polygons,  67. 

Polyhedral  angles,  277. 

Polyhedron,  287. 

Postulate,  8,  118. 

Premises,  7,  30. 

Prism,  3SS. 

Problem,    8,     118-132,     166-171, 

198-201,  230-238. 
Product  of  base  and  altitude,  181 
Projection  of  a  point  or  a  line,  19-1 , 

275. 

Proof,  7. 
Property,  72. 
Proportion,  137. 
Proposition,  8. 
Pure  number,  100. 
Pyramids,  5I09. 
Pythagorean  proposition,  190. 

Quadrant,  355. 
Quadrilateral,  68. 
Quantity,  99. 
Quotient,  100. 

Radius.  82,  211,  347. 
Ratio,  99,  100.  103,  137. 
Ratio  of  similitude,  153. 
Kr:ctangle,  69,  70. 


INDEX. 


395 


Rectangular  parallelepiped,  297. 

R-  ctilinear  figui 

Red  net  ii>  ad  absitrdiun,  5~>. 

Re-entrant  angl< 

Regular  polygons,  2(M 

Rt-gular  polyhedrons,  322. 

Regular  prisms,  '.'>'.' 

R. -pillar  pyramids,  309. 

Rhomb... (1,  Ml). 

Rhombus,  70. 

Right  angle.  5. 

Right  circular  cone,  310. 

Right  cylinder,  :;:',(). 

Right  parallelopiped,  297. 

Right  section  ot  prism,  288. 

Right  triangle,  18. 

Scalene  triangle,  18. 

Scholium,  8,  13!). 

Secant,  S3. 

Secant  line,  41. 

Sector,  S3. 

Segment,   83. 

Sides  of  an  angle,  5. 

Sides  of  a  polygon,  (57. 

Sides  of  a  triangle,  18. 

Similar  arcs,  etc.,  2MD 

Similar  polygons,   l.VJ 

Similitude,  1."):; 

Slant  height,  3(>D 

Small  circle  of  sphere,  .'!  ID 

Solid,  1. 

Solid  geometry,  4,  311. 

Special  enunciation,  7. 

Sphere,  347. 

Spherical  area,  371. 

Spherical  polygon,  3.~>D. 

Spherical  pyramid,  380. 

Spherical  sector,  378. 

Spherical  segment,  378. 


Spherical  triangle,  359. 

Squar- 

Square  of  a  line,  181,  190. 

Straight  angle,  .">. 

Straight  edge,  118. 

Straight  line,  3. 

Subtend,  83. 

Supplements,  7. 

Supplementary  adjacent  angles,  7. 

Surface,  1,  2,  347. 

Symbols  and  abbreviations,  9. 

Symmetrical    spherical    triangle. 

360. 
Symmetrical     trihedral     angles. 

277. 

Tangent,  83,  340,  347. 
Terms  of  a  proportion,  137. 
Tetrahedron.  287. 
Theorem,  7. 

Third  proportional,  168. 
Transversal,   11 , 
Trapezium,  C.S. 
Trapezoid,  OS. 
Triangle,  IS,  <>s 
Trihedral  angle,  2?  7. 
Trirectangular  triangle,  37 1. 
Truncated  pyramid,  310. 
Turn  through  an  angle,  4. 

Unit  of  measure,  90,  111,  I  7<>,  181. 

Variable,  104,  105. 

Vertex  of  an  angle,  5. 

Vertical  angles,  G. 

Vertical  angle  of  a  triangle,  18. 

Vertices  of  a  triangle,   IS. 

Vertices  of  a  polygon,  07. 

Zone,  378. 


YC  22379 


M306168 


V3 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


